4A 4B 4C 4D 4E 4F 4G 4H 4I 4J Trigonometry 4K...

Raylene competes in the sport of orienteering and is preparing for a race. The race is to complete a triangular course. The fi rst leg of the course is 1.2 km on a bearing of 200°T. The second leg is 2.3 km on a bearing of 320°T. Raylene then needs to calculate the distance and the bearing of the third leg, which will return her to the start/fi nish line. Problems involving bearings — used in orienteering as well as navigation and other practical areas — can be solved using trigonometry. In this chapter, we will further develop the trigonometric skills covered in Year 9 and apply them to many practical situations.

4A Pythagoras’ theorem 4B Pythagoras’ theorem in three dimensions 4C Trigonometric ratios 4D Using trigonometry to calculate

side lengths 4E Using trigonometry to calculate

angle size 4F Angles of elevation and depression 4G Bearings 4H The unit circle — quadrant 1 4I Circular functions 4J Graphs of trigonometric functions 4K Applications

4

Trigonometry

Chapter 4 Trigonometry 145

are yoU ready?

Try the questions below. If you have diffi culty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy.

Rounding to a given number of decimal places

1 Round the following numbers to 3 decimal places.a 0.6845 b 1.3996 c 0.7487

Rounding the size of an angle to the nearest minute and second

2 Round the following angles: i to the nearest minute ii to the nearest second.a 15°32′40.5″ b 63°15′32.4″ c 27°10′15.8″

Labelling the sides of a right-angled triangle

3 Label the sides of the following right-angled triangles using the letters H (for hypotenuse), O (for opposite) and A (for adjacent) with respect to angle θ.a b c

Rearranging formulas

4 Rearrange each of the following formulas to make x the subject.

a tan (15°) = x

30b tan (28°) = 4 2.4 2.4 2

xc

xtan(tan(ta )64°

= 5.3

Drawing a diagram from given directions

5 Draw a diagram for each of the following situations.a Kate’s bushwalking route took her from A to B, a distance of 5 km at a bearing of 25°T

then to C, a further distance of 7.5 km at a bearing of 120°T.b A ship steamed S20°E for a distance of 180 km, then the ship travelled N60°W for a

further 70 km.

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Digital doc SkillSHEET 4.1Rounding to a given number of decimal places

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Digital doc SkillSHEET 4.2Rounding the size of an angle to the nearest minute and second

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Digital doc SkillSHEET 4.3Labelling the sides of a right-angled triangle

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Digital doc SkillSHEET 4.5Rearranging formulas

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Digital doc SkillSHEET 4.6Drawing a diagram from given directions

146

pythagoras’ theoremintroductionIn Year 9, Pythagoras’ theorem and the three trigonometric ratios (sine, cosine and tangent) were introduced. Later in this chapter, the three ratios will be revisited and their various applications will be discussed. Also, the concept of the unit circle and circular functions will be introduced. Finally, we will investigate the graphs of sine and cosine functions. In this section, we will revisit Pythagoras’ theorem.

review of pythagoras’ theoremIn any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The rule is written as c2 = a2 + b2 where a and b are the two shorter sides and c is the hypotenuse.

The hypotenuse is the longest side of a right-angled triangle and isalways the side that is opposite the right angle.

Pythagoras’ theorem gives us a way of fi nding the length of the third side in a triangle, if we know the lengths of the two other sides.

Finding the hypotenuseWe are able to calculate the length of the hypotenuse when we are given the length of the two shorter sides by substituting into the formula c2 = a2 + b2.

It is easier to know what to do if you remember that:

fi nding a long side means addition.

For the triangle at right, calculate the length of the hypotenuse, x, correct to 1 decimal place.

Think WriTe/draW

1 Copy the diagram and label the sides a, b and c. Remember to label the hypotenuse as c.

c xa 4

b 7

2 Write Pythagoras’ theorem. c2 = a2 + b2

3 Substitute the values of a, b and c into this rule and simplify.

x2 = 42 + 72

= 16 + 49 = 65

4 Calculate x by taking the square root of 65. Round the answer correct to 1 decimal place.

x = 65 x = 8.1

4a

b

ac

4

7

x

Worked example 1

4

7

x

maths Quest 10 for Victoria for the Casio Classpad

147Chapter 4 Trigonometry

Finding a shorter sideSometimes a question will give you the length of the hypotenuse and ask you to find one of the shorter sides. In such examples, we need to rearrange Pythagoras’ formula. Given that c2 = a2 + b2, we can rewrite this as: a2 = c2 − b2

or b2 = c2 − a2.It is easier to know what to do in this case if you remember that:

finding a short side means subtraction.

Calculate the length, correct to 1 decimal place, of the unmarked side of the triangle at right.

Think WriTe/draW

1 Copy the diagram and label the sides a, b and c. Remember to label the hypotenuse as c; it does not matter which side is a and which side is b.

b 8

c 14

a

2 Write Pythagoras’ theorem for a shorter side. a2 = c2 − b2


a2 = 142 − 82

= 196 − 64 = 132

4 Find a by taking the square root of 132. Round to 1 decimal place.

a = 132 = 11.5 cm

In many cases we are able to use Pythagoras’ theorem to solve practical problems. We can model the problem by drawing a diagram, and use Pythagoras’ theorem to solve the right-angled triangle. We then use the result to give a worded answer.

A ladder that is 4.5 m long leans up against a vertical wall. The foot of the ladder is 1.2 m from the wall. How far up the wall does the ladder reach? Give your answer correct to 1 decimal place.

Think WriTe/draW

1 Draw a diagram and label the sides a, b and c. Remember to label the hypotenuse as c.

b 1.2 m

c 4.5 ma

Worked example 2

8 cm

14 cm

Worked example 3

148

2 Write Pythagoras’ theorem for a shorter side. a2 = c2 − b2


a2 = 4.52 − 1.22

= 20.25 − 1.44 = 18.81

4 Find a by taking the square root of 18.81. Round to 1 decimal place and include the unit of measurement (m).

a = 18 81. = 4.3 m

5 Answer the question in a sentence. The ladder will reach a height of 4.3 m up the wall.

Some questions will require you to decide which method is needed to solve the problem. A diagram will help you decide whether you are fi nding the hypotenuse or one of the shorter sides. Other questions will require you to fi nd an unknown on more than one length. In such cases, draw a labelled diagram, write the equation and solve for the unknown. The following worked example demonstrates this technique.

Calculate the value of the pronumeral, correct to 2 decimal places, in the triangle at right.

Think WriTe/draW

1 Copy the diagram and label the sides a, b and c. b 3x

a 2x

c 78

2 Write Pythagoras’ theorem. c2 = a2 + b2


782 = (3x)2 + (2x)2

6084= 9x2 + 4x2

6084= 13x2

4 Rearrange the equation so that the pronumeral is on the left-hand side of the equation.

13x2=6084

5 Divide both sides of the equation by 13. 13

13

2x =

608413

x2=468

6 Find x by taking the square root. Round the answer correct to 2 decimal places.

x = 468 = 21.63

Worked example 4

3x

2x

78



The hypotenuse is the longest side of the triangle and is opposite the right angle. 1. On your diagram, check whether you are fi nding the length of the hypotenuse or one of 2. the shorter sides. The length of the hypotenuse can be found if we are given the length of the two shorter 3. sides by using the formula c2 = a2 + b2.The length of the shorter side can be found if we are given the length of the hypotenuse 4. and the other shorter side by using the formula: a2 = c2 − b2 or b2 = c2 − a2. When using Pythagoras’ theorem, always check the units given for each measurement. 5. If necessary, convert all measurements to the same units before using the rule.6. Worded problems can be solved by drawing a diagram and using Pythagoras’ theorem to 7. solve the problem. Worded problems should be answered in a sentence.8.

rememBer

pythagoras’ theorem 1 We 1 For each of the following triangles, calculate the length of the hypotenuse, giving

answers correct to 2 decimal places.a 4.7

6.3

b

27.1

19.3 c

562

804

d

7.4

10.3

e 0.9

2.7

f

87

152

2 We2 Find the value of the pronumeral, correct to 2 decimal places.a

30.1

47.2

sb

1.98

2.56

t

c

8.4

17.52

u

d 0.28

0.67

v

e 2870

1920w

f

468

114

x

exerCise

4a

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Digital docSkillSHEET 4.1

Rounding to a given number

of decimal places

150

3 We3 The diagonal of the rectangular sign at right is 34 cm. If the height of this sign is 25 cm, find the width.

4 A right-angled triangle has a base of 4 cm and a height of 12 cm. Calculate the length of the hypotenuse to 2 decimal places.

5 Calculate the lengths of the diagonals of squares that have side lengths of:a 10 cm b 17 cm c 3.2 cm.

6 The diagonal of a rectangle is 120 cm. One side has a length of 70 cm. Determine:a the length of the other sideb the perimeter of the rectanglec the area of the rectangle.

7 We4 Find the value of the pronumeral, correct to 2 decimal places for each of the following.a

4x

x

25

b 3x 3x

18

c

6x

2x

30

8 An isosceles triangle has a base of 30 cm and a height of 10 cm. Calculate the length of the two equal sides.

9 An equilateral triangle has sides of length 20 cm. Find the height of the triangle.

10 A right-angled triangle has a height of 17.2 cm, and a base that is half the height. Calculate the length of the hypotenuse, correct to 2 decimal places.

11 The road sign shown below is in the form of an equilateral triangle. Find the height of the sign and, hence, find its area.

76 cm

12 A flagpole, 12 m high, is supported by three wires, attached from the top of the pole to the ground. Each wire is pegged into the ground 5 m from the pole. How much wire is needed to support the pole?

13 Ben’s dog ‘Macca’ has wandered onto a frozen pond, and is too frightened to walk back. Ben estimates that the dog is 3.5 m from the edge of the pond. He finds a plank, 4 m long, and thinks he can use it to rescue Macca. The pond is surrounded by a bank that is 1 m high. Ben uses the plank to make a ramp for Macca to walk up. Will he be able to rescue his dog?

14 Sarah goes canoeing in a large lake. She paddles 2.1 km to the north, then 3.8 km to the west. Use the triangle at right to find out how far she must then paddle to get back to her starting point in the shortest possible way.

Starting point

2.1 km

3.8 km



15 A baseball diamond is a square of side length 27 m. When a runner on first base tries to steal second base, the catcher has to throw the ball from home base to second base. How far is that throw?

27 m

Second base

Catcher

Firstbase

16 Penny, a carpenter, is building a roof for a new house. The roof has a gable end in the form of an isosceles triangle, with a base of 6 m and sloping sides of 7.5 m. She decides to put 5 evenly spaced vertical strips of wood as decoration on the gable as shown at right. How many metres of this decorative wood does she need?

17 Calculate the length, in mm, of the hypotenuse of a right-angled triangle, if the two shorter sides are 5 cm and 12 cm. Give your answer to 2 decimal places.

18 The hypotenuse and one other side of a right-angled triangle are given for each case below. Find the length of the third side in the units specified.a Sides 46 cm and 25 cm, third side in mmb Sides 843 mm and 1047 mm, third side in cmc Sides 4500 m and 3850 m, third side in kmd Sides 20.3 cm and 123 mm, third side in cme Sides 6420 mm and 8.4 m, third side in cmf Sides 0.358 km and 2640 m, third side in mg Sides 491 mm and 10.8 cm, third side in mmh Sides 379 000 m and 82 700 m, third side in km

19 A rectangle measures 35 mm by 4.2 cm. Calculate the length of its diagonal in millimetres to 2 decimal places.

20 A rectangular envelope has a length of 21 cm and a diagonal measuring 35 cm. Calculate:a the width of the envelopeb the area of the envelope.

21 A sheet of A4 paper measures 210 mm by 297 mm. Calculate the length of the diagonal in centimetres to 2 decimal places.

7.5 m

6 m

7.5 m

152

22 A right-angled triangle has a hypotenuse of 47.3 cm and one other side of 30.8 cm. Calculate the area of the triangle.

23 A swimming pool is 50 m by 25 m. Peter is bored by his usual training routine, and decides to swim the diagonal of the pool. How many diagonals must he swim to complete his normal distance of 1200 m? Give your answer to 2 decimal places.

24 Sarah is making a gate that has to be 1200 mm wide. It must be braced with a diagonal strut made of a different type of timber. She has only 2 m of this kind of timber available. What is the maximum height of the gate that she can make?

25 A hiker walks 4.5 km west, then 3.8 km south. How far in metres is she from her starting point? Give your answer to 2 decimal places.

26 A square has a diagonal of 10 cm. What is the length of each side?

27 Wally is installing a watering system in his garden. The pipe is to go all around the edge of the rectangular garden, and have a branch diagonally across the garden. The garden measures 5 m by 7.2 m. If the pipe costs $2.40 per metre (or part thereof), what will be the total cost of the pipe?

28 The size of a rectangular television screen is given by the length of its diagonal. What is the size of the screen below to the nearest centimetre if its dimensions are 158 cm wide and 96 cm deep?

pythagoras’ theorem in three dimensionsMany real-life situations involve 3-dimensional (3-D) shapes: shapes with length, width and height. Some common 3-D shapes used in this section include boxes, pyramids and right-angled wedges.

Box Pyramid Right-angled wedge

The important thing about 3-D shapes is that in a diagram, right angles may not look like right angles, so it is important to redraw sections of the diagram in two dimensions, where the right angles can be seen accurately.

4B



Determine the length AG in this box.

Think WriTe/draW

1 Draw the diagram in 3-D. A B

CF

GH

DE

10 cm

5 cm

6 cm

2 Draw, in 2-D, a right-angled triangle that contains AG and label the sides. Only 1 side is known, so we need to fi nd another right-angled triangle to use.

A

E G

6

3 Draw EFGH in 2-D and show the diagonal EG. Label the side EG as x. We have two of the three side lengths so we can calculate the unknown. x

E F

H G10

55

4 Use Pythagoras’ theorem to calculate EG. c2 = a2 + b2

x2 = 52 + 102

= 25 + 100 = 125

x = 125 = 11.18 cm

5 Place this information on triangle AEG. Label the side AG as y. Now we have two of the three side lengths. y

A

E G11.18

6

6 Use Pythagoras’ theorem to fi nd AG. c2 = a2 + b2

y2 = 62 + ( )( )125( )125( )2

= 36 + 125 = 161 y = 161 = 12.69

7 Answer the question in a sentence. The length of AG is 12.69 cm.

Worked example 5

A B

CF

GH

DE

10 cm

5 cm

6 cm

154

A piece of cheese in the shape of a right-angled wedge sits on a table. It has a rectangular base measuring 14 cm by 8 cm, and is 4 cm high at the thickest point. An ant crawls diagonally across the sloping face. How far, to the nearest millimetre, does the ant walk?

Think WriTe/draW

1 Draw a diagram in 3-D and label the vertices.Mark BD, the path taken by the ant, with a dotted line.

4 cm

8 cm14 cm

x

A

B C

D

E F

2 Draw in 2-D a right-angled triangle that contains BD, and label the sides. Only one side is known, so we need to fi nd another right-angled triangle to use.

4

B

ED

3 Draw EFDA in 2-D, and show the diagonal ED. Label the side ED as x.

x

E F

A D14

88

4 Use Pythagoras’ theorem to calculate ED. c2 = a2 + b2

x2 = 82 + 142

= 64 + 196 = 260 x = 260 = 16.12 cm

5 Place this information on triangle BED. Label the side BD as y.

4

16.12

B

ED

y

6 Solve this triangle for BD. c2 = a2 + b2

y2 = 42 + 2602( )

= 16 + 260 = 276 y = 276 = 16.61 cm

7 Check the answer’s units. We need to convert cm to mm, so multiply by 10.

= 166.1 mm

8 Answer the question in a sentence. The ant walks 166 mm, correct to the nearest millimetre.

Worked example 6



Pythagoras’ theorem can be used to solve problems in three dimensions (3-D).1. Some common 3-D shapes include boxes, pyramids and right-angled wedges.2. To solve problems in 3-D it is helpful to draw sections of the original shape in 3. two dimensions (2-D).

rememBer

pythagoras’ theorem in three dimensionsWhere appropriate in this exercise, give answers correct to 2 decimal places.

1 We5 Calculate the length, AG.a A B

C

F

GH

D

E

10

10

10

b A B

C

F

GH

D

E

10

5

5

c A B

C

F

GH

D

E

8.2

7.3

10.4

2 We6 Calculate the length of CE in the wedge at right and, hence, obtain AC.

3 If DC = 3.2 m, AC = 5.8 m, and CF = 4.5 m in the figure at right, calculate the length of AD and BF.

4 Calculate the length of BD and, hence, the height of the pyramid at right.

5 The pyramid ABCDE has a square base. The pyramid is 20 cm high. Each sloping edge measures 30 cm. Calculate the length of the sides of the base.

exerCise

4B

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Drawing 3-D shapes

4

710D

A B

CF

E

D

A B

CF

E

V

A B

DC

8

8

8

E

A

M

B

D C

EM = 20 cm

156

6 The sloping side of a cone is 10 cm and the height is 8 cm. What is the length of the radius of the base?

7 An ice-cream cone has a diameter across the top of 6 cm, and sloping side of 13 cm. How deep is the cone?

8 A piece of cheese in the shape of a right-angled wedge sits on a table. It has a base measuring 20 mm by 10 mm, and is 4 mm high at the thickest point, as shown in the figure. A fly crawls diagonally across the sloping face. How far, to the nearest millimetre, does the fly walk?

9 Jodie travels to Bolivia, taking with her a suitcase as shown in the photo. She buys a carved walking stick 1.2 m long. Will she be able to fit it in her suitcase for the flight home?

65 cm

90 cm

30 cm

10 A desk tidy is shaped like a cylinder, height 18 cm and diameter 10 cm. Pencils that are 24 cm long rest inside. What lengths of the pencils are above the top of the cylinder?

11 A 10-m high flagpole is in the corner of a rectangular park that measures 240 m by 150 m. a Calculate: i the length of the diagonal of the park ii the distance from A to the top of the pole iii the distance from B to the top of the pole.b A bird fl ies from the top of the pole to the centre of the park. How far does it fl y?

12 A candlestick is in the shape of two cones, joined at the vertices as shown. The smaller cone has a diameter and sloping side of 7 cm, and the larger one has a diameter and sloping side of 10 cm. How tall is the candlestick?

8 cm 10 cm

r

A

B C

DF

20 mm10 mm

4 mmE

A

B

240 m

150 m

10 m



13 The total height of the shape at right is 15 cm. Calculate the length of the sloping side of the pyramid.

14 A sandcastle is in the shape of a truncated cone as shown. Calculate the length of the diameter of the base.

15 A tent is in the shape of a triangular prism, with a height of 120 cm as shown at right. The width across the base of the door is 1 m, and the tent is 2.3 m long. Calculate the length of each sloping side, in metres. Then calculate the area of fabric used in the construction of the sloping rectangles which form the sides.

Trigonometric ratiosangles and the calculatorLast year you were shown that each angle has specifi c values for its sine, cosine and tangent. These values are needed for practically every trigonometry problem and can be obtained with the aid of a calculator.

Calculate the value of each of the following, correct to 4 decimal places, using a CAS calculator.a cos (65°57′) b tan (56°45′30″)

Think WriTe/display

1

a 2 Write your answer to the correct number of decimal places.

a cos (65°57′) = 0.4075

b 2 Write your answer to the correct number of decimal places.

b tan (56°45′30″) = 1.5257

15 cm

14 cm14 cm

6 cm

30 cm 32 cm

20 cm

120 cm

1 m2.3 m

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Digital docWorkSHEET 4.1Pythagoras’

theorem

4C

Worked example 7

On the Main screen, use the soft keyboard to access trigonometric expressions. For each entry, insert cos or tan and then tap:• Action• Transformation• dmsComplete the entry lines as:• cos(dms(65,57)• tan(dms(56,45,30)Press E after each entry.

158

Calculate the size of angle θ, correct to the nearest degree, given sin (θ) = 0.6583.

Think WriTe/display

1 Write the given information.2 To find the size of the angle, we need to

‘undo’ sine with its inverse, sin−1.

3

4 Write your answer to the nearest degree. θ = 41°

We also need to be able to find an angle correct to either the nearest minute or nearest second. When we use an inverse trigonometric function, the angle is expressed in degrees as a decimal.

Calculate the value of θ:a correct to the nearest minute, given that cos (θ) = 0.2547b correct to the nearest second, given that tan (θ) = 2.364.

Think WriTe/display

a 1 a

2 Write your answer, rounding to the nearest minute. Remember there are 60 minutes in 1 degree and 60 seconds in 1 minute. Hence, for the nearest minute, we round up at 30″ or higher.

cos (0.2547) = 75°15′

Worked example 8

Worked example 9


On the Main screen, use the soft keyboard to access trigonometric expressions. Complete the entry line as:

sin−1(0.6583)

Then press E.


cos−1(0.2547)

Press E. Then tap:• Action• Transformation• toDMS• D• E


b 1 b

2 Write your answer, rounding to the nearest second.

tan (2.364) = 67°4′16″

Review of SOH CAH TOAFrom our work in Year 9 we discovered that we were able to fi nd a side length in a right-angled triangle if we were given one other side length and the size of one of the acute angles. These sides and angle were related using one of the three trigonometric ratios.

The sine ratioThe sine ratio is defi ned as the ratio of the length of the side opposite angle θ (O) to the length of the hypotenuse (H). This is

written as sin (θ ) = OH

.

The cosine ratioThe cosine ratio is defi ned as the ratio of the length of the adjacent side (A) to the length of the hypotenuse (H) and is written as

cos (θ ) = AH

.

The tangent ratioThe tangent ratio is defi ned as tan (θ ) = O

A, where O is the length

of the side opposite angle θ and A is the length of the side adjacent

to it.

Having defi ned the three trigonometric ratios, we need to decide in each case which of the three to use. We do this by labelling the sides relative to the angle we have been given. We then select the ratio that contains both the side we are fi nding and the side we have been given.

Hypote

nuse

θ

Opposite

Hypote

nuse

θ

Adjacent

θ

Adjacent

Opposite


tan−1(2.364)

Press E. Then tap:• Action• Transformation• toDMS• D• E

160

The three ratios can be remembered easily by using the mnemonic or abbreviationSOH CAH TOA: • SOH stands for ‘Sine, Opposite, Hypotenuse’.• CAH stands for ‘Cosine, Adjacent, Hypotenuse’. • TOA stands for ‘Tangent, Opposite, Adjacent’.

For this triangle, write the expressions for the sine, cosine and tangent ratios of the given angle.

θ

ac

b

THINK WRITE/DRAW

1 Label the diagram using the symbols O, A, H with respect to the given angle (angle θ).

θ

a = O

c = H

b = A

2 From the diagram, identify the values of O (opposite side), A (adjacent side) and H (the hypotenuse).

O = a, A = b, H = c

3 Write the formula for each of the sine, cosine and tangent ratios.

sin (θ ) = OH

, cos (θ ) = AH

, tan (θ ) = OA

4 Substitute the values of A, O and H into each formula.

sin (θ ) = ac

, cos (θ ) = bc

, tan (θ ) = ab

Write the trigonometric ratio which must be used in order to fi nd the value of the pronumeral in each of the following triangles.

a

615

b

b

50°

18

x

WORKED EXAMPLE 10

WORKED EXAMPLE 11

Maths Quest 10 for Victoria for the Casio ClassPad


THINK WRITE/DRAW

a 1 Label the sides of the triangle whose lengths are given, using the appropriate symbols.

a

6 = O15 = H

b

2 We are given the lengths of the opposite side (O) and the hypotenuse (H). Write the ratio that contains both of these sides.

sin (θ ) = OH

3 Identify the values of the pronumerals. O = 6, H = 15

4 Substitute the values of the pronumerals into the ratio. (Since the given angle is denoted with the letter b, replace θ with b.)

sin (b) = 6

15

b 1 Label the sides of the triangle whose lengths are either given, or need to be found, using the appropriate symbols.

b

50°

18 = Ax = O

2 The length of the adjacent side (A) is given and the length of the opposite side (O) needs to be found. Write the ratio that contains these sides.

tan (θ ) = OA

3 Identify the values of the pronumerals. O = x, A = 18, θ = 50°

4 Substitute the values of the pronumerals into the ratio. tan (50°) = x

18

When using the calculator to fi nd values of sine, cosine and tangent, make sure the 1. calculator is in Degree mode.To fi nd the size of an angle whose sine, cosine or tangent is given, perform an inverse 2. operation; that is, sin

−1, cos−1 or tan

−1.Use the calculator’s conversion function to convert between decimal degrees and 3. degrees, minutes and seconds.There are 60 minutes in 1 degree and 60 seconds in 1 minute.4. The three trigonometric ratios, sine, cosine and tangent, are defi ned as:5.

sin (θ ) = OH

, cos (θ ) = AH

and tan (θ ) = OA

,

where H is the hypotenuse, O is the opposite side and A is the adjacent side.The three ratios are abbreviated to the useful mnemonic SOH CAH TOA.6. To determine which trigonometric ratio to use, follow these steps.7. (a) Label the sides of the right-angled triangle that are either given, or need to be

found, using the symbols O, A, H with respect to the angle in question.(b) Consider the sides that are involved and write the trigonometric ratio containing

both of these sides. (Use SOH CAH TOA to assist you.)(c) Identify the values of the pronumerals in the ratio.(d) Substitute the given values into the ratio.

REMEMBER

162

Trigonometric ratios 1 Calculate each of the following, correct to 4 decimal places.

a sin (30°) b cos (45°) c tan (25°)d sin (57°) e tan (83°) f cos (44°)

2 We7 Calculate each of the following, correct to 4 decimal places.a sin (40°30′) b cos (53°57′) c tan (27°34′)d tan (123°40′) e sin (92°32′) f sin (42°8′)g cos (35°42′35″) h tan (27°42′50″) i cos (143°25′23″)j sin (23°58′21″) k cos (8°54′2″) l sin (286°)m tan (420°) n cos (845°) o sin (367°35′)

3 We8 Find the size of angle θ, correct to the nearest degree, for each of the following.a sin (θ) = 0.763 b cos (θ) = 0.912 c tan (θ)= 1.351d cos (θ) = 0.321 e tan (θ) = 12.86 f cos (θ) = 0.756

4 We9a Find the size of the angle θ in each of the following, correct to the nearest minute.a sin (θ) = 0.814 b sin (θ) = 0.110 c tan (θ) = 0.015d cos (θ) = 0.296 e tan (θ) = 0.993 f sin (θ) = 0.450

5 We9b Find the size of the angle θ in each of the following, correct to the nearest second.a tan (θ) = 0.5 b cos (θ) = 0.438 c sin (θ) = 0.9047d tan (θ) = 1.1141 e cos (θ) = 0.8 f tan (θ) = 43.76

6 Find the value for each of the following, correct to 3 decimal places.

a 3.8 cos (42°) b 118 sin (37°) c 2.5 tan (83°)

d 245sin ( )°

e 22014cos ( )°

f 2 235 18co2 2co2 2s (2 2s (2 2 )

5 1si5 1n (5 1n (5 1 )

°°

g 12 860 32

.tan (tan (ta )° ′

h 18 735 25 42

.sin ( )° ′ ′42′ ′42 ′

i 55 789 21

.cos ( )° ′

j 3 8 1 54 5 25 45. t3 8. t3 8 an ( )1 5( )1 51 4( )1 44( )4

. s4 5. s4 5 in. sin. s ( )25( )25 45( )451 5°1 51 5( )1 5°1 5( )1 5 ′ ′( )′ ′( )1 4( )1 4′ ′1 4( )1 44( )4′ ′4( )4 ′( )′( )

°( )°( )′( )′( )k 2 5

10 4 8. s2 5. s2 5 in. sin. s ( )27( )27 8( )8. c4 8. c4 84 8os4 8( )4 8( )4 83 2( )3 2

°( )°( )′( )′( )°3 2( )3 2°3 2( )3 2′( )′( )

l 3 2 520 8. c3 2. c3 2 os ( )34( )34 52( )52. s0 8. s0 8 in. sin. s ( )12( )12 48( )48

°( )°( )′( )′( )°( )°( )′( )′( )

7 We10 For each of the following triangles, write the expressions for ratios of each of the

given angles: i sine ii cosine iii tangenta d

f e

b

i h

g

c

jk

l

d o

n

m

e

a b

c

f

v u

t

exerCise

4C

eBookpluseBookplus


Labelling the sides of a

right-angled triangle



8 We 11 Write the trigonometric ratio which must be used in order to find the value of the

pronumeral in each of the following triangles.

a

1215

b 25

30

c 5

4

d p

2.7 e t

17

35

f 14.3

17.5

g

x

7

15

h

20 31

i

9.8

3.1

9 Consider the right-angled triangle shown at right.a Label each of the sides using the letters O, A, H with

respect to the 41° angle.b Measure the side lengths (to the nearest millimetre).c Determine the value of each trigonometric ratio. (Where

applicable, answers should be given correct to 2 decimal places.)

i sin (41°) ii cos (41°) iii tan (41°)d What is the value of the unknown angle, α?e Determine the value of each of these trigonometric ratios, correct to 2 decimal places. i sin (α) ii cos (α) iii tan (α) (Hint: First relabel the sides of the triangle with respect to angle α.)f What do you notice about the relationship between sin (41°)and cos (α)?g What do you notice about the relationship between sin (α) and cos (41°)?h Make a general statement about the two angles.

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Selecting an appropriate

trigonometric ratio based

on the given information

41

164

Using trigonometry to calculate side lengths Consider this right-angled triangle.

Labelling the sides with respect to the 42° angle, we can see that the unknown side is opposite and we are given the hypotenuse.

Using our calculator, we know that the sine ratio of a 42° angle is approximately 0.6691.

From the diagram at right, sin (42°) = x

24. We can now solve

this equation to fi nd the value of x.We are therefore able to calculate a side length if we are

given the size of an angle and one other side.The solution to the above problem is:

sin (θ) = OH

the sine ratio formula

sin (42°) = x

24 the result of substituting into the formula

x = 24 × sin (42°) rearranging the formula to make x the subject

x ≈ 16.06 m the result of the calculation.We need to apply this method using any of the three trigonometric ratios to fi nd a side length.

The steps used in solving the problem are as follows.Step 1. Label the sides of the triangle, which are either given, or need to be found, with respect

to the given angle.Step 2. Consider the sides involved and determine which of the trigonometric ratios is required.

(Use the mnemonic SOH CAH TOA to help you.) (a) Use the sine ratio if the hypotenuse (H) and the opposite side (O) are involved. (b) Use the cosine ratio if the hypotenuse (H) and the adjacent side (A) are involved. (c) Use the tangent ratio if the opposite (O) and the adjacent (A) sides are involved.Step 3. Substitute the values of the pronumerals into the ratio.Step 4. Solve the resultant equation for the unknown side length.

Find the value of the pronumeral for each of the following. Give answers correct to 3 decimal places.a

35

6 cma

b

32

0.346 cm f

Think WriTe/draW

a 1 Label the sides of the triangle, which are either given, or need to be found.

a

35

6 cma

H O

4d

42

eBookpluseBookplus

Interactivityint-1146Using

trigonometry

24 m

42

x

24 m

42

x

Opp

osite

Hypotenuse

Adjacent

Worked example 12



2 Identify the appropriate trigonometric ratio to use.

sin (θ)= OH

3 Substitute O = a, H = 6, θ = 35°. sin (35°)= a6

6 sin (35°)=a4 Make a the subject of the equation. a = 6 sin (35°)

a ≈ 3.441 cm5 Calculate and round the answer, correct to 3 decimal places.

b 1 Label the sides of the triangle, which are either given, or need to be found.

b

32

0.346 cm f

H A


cos (θ)= AH

3 Substitute A = f, H = 0.346 and θ = 32°. cos (32°)= f

0.346

0.346 cos (32°)=f4 Make f the subject of the equation. f = 0.346 cos (32°)

≈ 0.293 cm5 Calculate and round the answer, correct to 3 decimal places.

In Worked example 12, the unknown side was in the numerator of the fraction when we substituted the given information. This calculation step will differ when the unknown value is in the denominator.

Find the value of the pronumeral in the triangle shown. Give the answer correct to 2 decimal places.

5120 m

P

Think WriTe/draW

1 Label the sides of the triangle, which are either given, or need to be found.

5120 m

PA

H O

2 Identify the appropriate trigonometric ratio to use. tan (θ)= OA

3 Substitute O = 120, A = P and θ = 5°. tan (5°)= 120P

Worked example 13

166

4 Make P the subject of the equation. (i) Multiply both sides of the equation by P.

(ii) Divide both sides of the equation by tan (5°).

P × tan (5°)=120

P = 120

5tan (tan (ta )°

5 Calculate and round the answer, correct to 2 decimal places.

P ≈ 1371.61 m

The trigonometric ratios can be used to fi nd a side length in a right-angled triangle 1. when we are given one other side length and one of the acute angles.The calculation step will differ depending upon whether the unknown is in the 2. numerator or denominator of the equation formed after substitution.

rememBer

Using trigonometry to calculate side lengths 1 We12 Find the length of the unknown side in each of the following, correct to 3 decimal

places.

a

60

10 cm a

b

25

8

a

c

3114

x

2 We13 Find the length of the unknown side in each of the following triangles, correct to 2 decimal places.

a 71

2.3 m

mb 13

4.6 m

n

c

68

94 mm

t

exerCise

4d



3 Find the length of the unknown side in each of the following, correct to 2 decimal places.

a

40 26'

43.95 m

x

b 8 52' 45''

11.7 m

P c

18 12

'

14 m

t

d

34 42

'11.2 mm

x

e

75.23 km

x

2125

' 34"

f

80.9 cmx

6 25'

4 Find the value of the pronumeral in each of the following, correct to 2 decimal places.

a

43.9 cm

x

46

b 23.7 m

y

36 42'

c

12.3 mz

34 12'

d

15.3 mp

13 12'

e

47.385 km

pq

63 11'

f 0.732 km

b

a73 5'

5 Given that the angle θ is 42° and the length of the hypotenuse is 8.95 m in a right-angled triangle, find the length of:a the opposite side b the adjacent side.Give each answer correct to 1 decimal point.

6 A ladder rests against a wall. If the angle between the ladder and the ground is 35°and the foot of the ladder is 1.5 m from the wall, how high up the wall does the ladder reach?

168

Using trigonometry to calculate angle sizeTo fi nd the size of an angle using the trigonometric ratios, we need to be given the length of two sides.

For each of the following, fi nd the size of the angle, θ, correct to the nearest degree.a

5 cm3.5 cm

b

11 m

5 m

Think WriTe/draW


a

3.5 cm5 cm

H O

2 Identify the appropriate trigonometric ratio to use. We are given O and H, so choose the sine ratio.

sin (θ) = OH

3 Substitute O = 3.5 and H = 5 and evaluate the expression.

sin (θ) = 3.55

= 0.7

4 Make θ the subject of the equation using inverse sine.

θ = sin−1(0.7)

= 44.427 004°

5 Evaluate θ and round the answer, correct to the nearest degree.

θ ≈ 44°


b

5 m

11 m

O

A

2 Identify the appropriate trigonometric ratio to use. We are given O and A, so choose the tangent ratio.

tan (θ)=OA

4e

Worked example 14



3 Substitute O = 5 and A = 11. As the value of tan (θ) is a simple fraction, we do not need to evaluate the expression.

tan (θ) = 511

4 Make θ the subject of the equation using inverse tangent.

θ = tan−1 5

11

= 24.443 954 78°5 Evaluate θ and round the answer, correct

to the nearest degree. θ ≈ 24°

When asked for a more accurate measurement of an angle, we are able to use the calculator to fi nd an angle correct to the nearest minute or nearest second.

Find the size of angle θ in each of the triangles shown below.a 3.1 m

7.2 m

b

55 cm

42 cm

(Answer correct to (Answer correct to the nearest minute.) the nearest second.)

Think WriTe/draW


a 3.1 m

7.2 m

O

A


tan (θ)= OA

3 Substitute O = 7.2 and A = 3.1 and evaluate the expression.

tan (θ) = 7.23.1

= 2.322 580 645

4 Make θ the subject of the equation using inverse tangent.

θ = tan−1 (2.322 580 645)

5 Evaluate θ and write the calculator display. θ = 66.705 436 75°6 Use the calculator to convert the answer to

degrees, minutes and seconds and round the answer to the nearest minute.

=66°42′19.572″ θ ≈ 66°42′

Worked example 15

170


b

55 cm

42 cm

H

A


cos (θ)= AH

3 Substitute A = 42 and H = 55. cos (θ) = 4255

4 Make θ the subject of the equation using inverse cosine.

θ = cos−1 42

55

5 Evaluate θ and write the calculator display. θ = 40.214 171 02°6 Use the calculator to convert the answer

to degrees, minutes and seconds and round the answer to the nearest second.

=40°12′51.016″ θ ≈ 40°12′51″

The trigonometric ratios can be used to fi nd the size of the acute angles in a 1. right-angled triangle when we are given the length of two sides.To fi nd an angle size we need to use the inverse trigonometric functions.2. Answers may be given correct to the nearest degree, minute or second.3.

rememBer

Using trigonometry to calculate angle size 1 We14 Find the size of the angle, θ, in each of the following. Give your answer correct to the

nearest degree.a

4.85.2

b

3.2

4.7

c

3

8

2 We 15a Find the size of the angle marked with the pronumeral in each of the following. Give your answer correct to the nearest minute.a

12

17

b

4 m

7.2 m c

10

12

exerCise

4e

eBookpluseBookplus


Rounding angles to the

nearest degree



3 We 15b Find the size of the angle marked with the pronumeral in each of the following. Give your answer correct to the nearest second.

a

2

8

b

3 m

5 m

c

3.5

2.7

4 Find the size of the angle marked with the pronumeral in each of the following, giving your answer correct to the nearest degree.

a

15.3

13.5 a

b

77.3

89.4

c

c

92.7

106.4

b

d

e f

5 Find the size of each of the angles in the following, giving your answers correct to the nearest minute.

a

56.3

a

b27.2

b

0.798

d

e0.342

c

5.7x

y

2.3

18.743.7

d

12.36

13.85

18.56

e 12.2 cm7.3 cm

9.8 cm

eBookpluseBookplus

Digital docWorkSHEET 4.2

Using trigonometry

172

angles of elevation and depressionMany people use trigonometry at work. It is particularly important in careers such as the building trades, surveying, architecture and engineering. Trigonometric ratios have a variety of applications, some of which will be discussed in this section.

Trigonometric ratios can be used to solve problems. When solving a problem, the following steps can be of assistance.1. Sketch a diagram to represent the situation described in the problem.2. Label the sides of the right-angled triangle with respect to the angle involved.3. Identify what is given and what needs to be found. 4. Select an appropriate trigonometric ratio and use it to fi nd the unknown measurement.5. Interpret your result by writing a worded answer.

angles of elevation and depressionWhen we need to look up or down in order to see a certain object, our line of vision (that is, the straight line from the observer’s eye to the object) is inclined. The angle of inclination of the line of vision to the horizontal when looking up is referred to as the angle of elevation, and when looking down it is referred to as the angle of depression.

The angle of elevation is measured up from the horizontal line to the line of vision.

Angle ofelevation

Horizontal

The angle of depression is measured down from the horizontal line to the line of vision.

Angle of depression

Horizontal

For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of depression of A as seen from B.

A

Angle of elevationof B from A

Angle of depressionof A from B

B

4FeBookpluseBookplus

eLessoneles-0173

Height of a satellite



From an observer, the angle of elevation of the top of a tree is 50°. If the observer is 8 metres from the tree, fi nd the height of the tree.

Think WriTe/draW

1 Sketch a diagram and label the sides of the triangle with respect to the given angle. Let the height of the tree be h.

8 m50

h

O

A

2 Identify the appropriate trigonometric ratio. We are given A and need to fi nd O, so choose the tangent ratio.

tan (θ)= OA

3 Substitute O = h, A = 8 and θ = 50°. tan (50°) =h8

4 Rearrange to make h the subject. h = 8 tan (50°)5 Calculate and round the answer to 2 decimal places. ≈ 9.53

6 Give a worded answer. The height of the tree is 9.53 m.

To solve a problem involving trigonometric ratios, follow these steps:1. (a) Draw a diagram to represent the situation.(b) Label the diagram with respect to the angle involved (either given or that needs to

be found).(c) Identify what is given and what needs to be found.(d) Select an appropriate trigonometric ratio and use it to fi nd the unknown side or angle.(e) Interpret the result by writing a worded answer.The angle of elevation is measured up and the angle of depression is measured down 2. from the horizontal line to the line of vision.

HorizontalAngle ofdepression

Angle ofelevation

Horizontal

3. For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of depression of A as seen from B.

A



B

rememBer

Worked example 16

174

angles of elevation and depression 1 We 16 The angle of elevation from an observer to the top of a tree is 54°22′. If the tree is

known to be 12.19 m high, how far is the observer from it?

2 From the top of a cliff 112 m high, the angle of depression to a boat is 9°15′. How far is the boat from the foot of the cliff?

3 A person on a ship observes a lighthouse on the cliff, which is 830 metres away from the ship. The angle of elevation of the top of the lighthouse is 12°.a How far above sea level is the top of the lighthouse?b If the height of the lighthouse is 24 m, how high is the cliff?

4 At a certain time of the day a post, 4 m tall, casts a shadow of 1.8 m. What is the angle of elevation of the sun at that time?

5 An observer, who is standing 47 m from a building, measures the angle of elevation of the top of the building as 17°. If the observer’s eye is 167 cm from the ground, what is the height of the building?

6 A surveyor needs to determine the height of a building. She measures the angle of elevation of the top of the building from two points, 38 m apart. The surveyor’s eye level is 180 cm above the ground.a Find two expressions for the height of the

building, h, in terms of x using the two angles.b Solve for x by equating the two expressions

obtained in a.c Find the height of the building.

7 The height of another building needs to be determined but cannot be found directly. The surveyor decides to measure the angle of elevation of the top of the building from different sites, which are 75 m apart. The surveyor’s eye level is 189 cm above the ground.a Find two expressions for the height of the

building, h, in terms of x using the two angles.b Solve for x.c Find the height of the building.

8 A lookout tower has been erected on top of a cliff. At a distance of 5.8 km from the foot of the cliff, the angle of elevation to the base of the tower is 15.7° and to the observation deck at the top of the tower is 16° respectively as shown in the figure below. How high from the top of the cliff is the observation deck?

5.8 km15.7

16

exerCise

4F

eBookpluseBookplus


Drawing a diagram

from given directions

x

h

35 50'47 12'180 cm38 m

x

h

32 18'43 35'

189 cm75 m



9 Elena and Sonja were on a camping trip to the Grampians, where they spent their first day hiking. They first walked 1.5 km along a path inclined at an angle of 10° to the horizontal. Then they had to follow another path, which was at an angle of 20° to the horizontal. They walked along this path for 1.3 km, which brought them to the edge of the cliff. Here Elena spotted a large gum tree 1.4 km away. If the gum tree is 150 m high, what is the angle of depression from the top of the cliff to the top of the gum tree?

10 a Find the height of a telegraph pole in the photograph at right if the angle of elevation to the top of the pole is 8° from a point at the ground level 60 m from the base of the pole.

b Find the height of the light pole in the figure below.

43.3

60 m

11 From a point on top of a cliff, two boats are observed. If the angles of depression are 58° and 32° and the cliff is 46 m above sea level, how far apart are the boats?

12 Joseph is asked to obtain an estimate of the height of his house using any mathematical technique. He decides to use an inclinometer and basic trigonometry. Using the inclinometer, Joseph determines the angle of elevation, θ, from his eye level to the top of his house to be 42°. The point from which Joseph measures the angle of elevation is 15 m away from his house and the distance from Joseph’s eyes to the ground is 1.76 m.a Fill in the given information on the diagram provided

(substitute values for the pronumerals).b Determine the height of Joseph’s house.

13 The competitors of a cross-country run are nearing the finish line. From a lookout 100 m above the track, the angles of depression to the two leaders, Nathan and Rachel, are 40° and 62° respectively. How far apart are the two competitors?

150 m

1.4 km

20

101.5 km

1.3 km Angle of depression

8°

60 m

3258

46 m

x

dh

4062

100 m

176

14 A 2.05 m tall man, standing in front of a street light 3.08 m high, casts a 1.5 m shadow.a What is the angle of elevation from the ground to

the source of light?b How far is the man from the bottom of the light

pole?

BearingsIn Year 9, bearings were introduced as a method of navigation. These can be expressed as either a compass bearing or a true bearing.

Compass bearingsCompass bearings (also known as conventional bearings) are measured from the north–south line in either a clockwise or anticlockwise direction.

To identify the compass bearing of an object we need to state:1. whether the angle is measured from north (N) or south (S)2. the size of the angle3. whether the angle is measured in the direction of west (W) or east (E).

For example, the bearing of S20°E means the direction that is 20° from south towards east, while the bearing N40°W means the direction that is 40° from north towards west.

N

S

W E

S20 E

20

N

S

W E

N40 W

40

True bearingsTrue bearings are measured from north in a clockwise direction. They are always expressed as 3 digits.

The diagrams below show the bearings of 025° true and 250° true respectively. (These true bearings are more commonly written as 025°T and 250°T.)

N

S

W E

025 true

25

N

S

W E

250 T

250

eBookpluseBookplus


Elevation and depression

3.08 m 2.05 m

1.5 m

4G



A boat travels a distance of 5 km from P to Q in a direction of 035°T. a How far east of P is Q?b How far north of P is Q?c What is the true bearing of P from Q?

Think WriTe/draW

a 1 Draw a diagram to represent the situation. Label the hypotenuse and the opposite and adjacent sides.

a

5 km

x

y

N

35

P

A H

O

Q

2 To determine how far Q is east of P, we need to fi nd the value of x. We are given the length of the hypotenuse (H) and need to fi nd the length of the opposite side (O). Choose the sine ratio.

sin (θ)=OH

3 Substitute O = x, H = 5 and θ = 35°. sin (35°) = x5

4 Make x the subject of the equation. x = 5 sin (35°)5 Evaluate and round the answer, correct to

2 decimal places.= 2.87

6 Write the answer in words. Point Q is 2.87 km east of P.

b 1 To determine how far Q is north of P, we need to fi nd the value of y. This can be done in several ways, namely: using the cosine ratio, the tangent ratio, or Pythagoras’ theorem. Let’s use the cosine ratio.

b cos (θ)= AH

2 Substitute P = y, H = 5 and θ = 35°. cos (35°) = y5

3 Make y the subject of the equation. y = 5 cos (35°)4 Evaluate and round the answer, correct to

2 decimal places. = 4.10

5 Write the answer in words. Point B is 4.10 km north of A.

c 1 To fi nd the bearing of P from Q, we need to draw the compass directions through Q and then measure the angle in the clockwise direction from the north line through Q to the line PQ. Show the required angle on the diagram.

c N

35

N

P

Q

Worked example 17

178

2 Study the diagram. The angle that represents the true bearing is the sum of 180° (from north to south) and the angle, labelled α. Now the north lines through P and Q are parallel and so the line PQ is a transversal. Therefore angle 35° and angle α are equal (being alternate angles). Calculate the true bearing.

True bearing = 180° + αα = 35°

True bearing = 180° + 35°= 215°

3 Write the answer in words. The bearing of P from Q is 215°T.

Sometimes a person or an object (for example, a ship) changes direction during their journey. (This can even happen more than once.) In situations like this we are usually interested in the total distance the object has moved and its fi nal bearing from the starting point. The following worked example shows how to deal with such situations.

A boy walks 2 km on a true bearing of 090° and then 3 km on a true bearing of 130°. a How far east of the starting point is the boy at the completion of his walk? (Answer correct to

1 decimal place.)b How far south of the starting point is the boy at the completion of his walk? (Answer correct to

1 decimal place.)c What is the bearing of the boy (from the starting point), in degrees and minutes, at the completion

of his walk?

Think WriTe/draW

Draw a diagram of the boy’s journey.

2 km

3 km

N

130

N

a 1 The fi rst leg of the journey is due east so we fi nd the eastern component of the second leg. Construct a triangle about the second leg of the journey. We can calculate one of the missing angles by using the rule of supplementary angles: 180° − 130° = 50°.

a

2 km

A

O

H3 km

N

E130

50

N

y

x

2 We need to fi nd the eastern component of the journey, x, which is the opposite side and have been given the hypotenuse. Choose the sine ratio.

sin (θ)=OH

3 Substitute O = x, H = 3 and θ = 50°. sin (50°) =x3

4 Make x the subject of the equation. x = 3 sin (50°)

Worked example 18



5 Evaluate and round correct to 1 decimal place.

= 2.3 kmTotal distance east = 2 + 2.3

= 4.3 km

6 Add to this the 2 km east that was walked in the fi rst leg of the journey and give a worded answer.

The boy walked a total of 4.3 km east of the starting point.

b 1 In the fi rst part of the journey the boy has not moved south at all. Thus the distance that he moved south of the starting point is the southern component of the second leg, labelled y. (See the diagram in part a .)

b Distance south = y km

2 To fi nd y we can use Pythagoras’ theorem, as we know the lengths of two out of three sides in the right-angled triangle. Note that the hypotenuse, c, is 3 and one of the sides is 2.3, as found in part a . Round the answer correct to 1 decimal place.Note: Alternatively, the cosine ratio could have been used.

a2 = c2 − b2 cos (50°)=y3

y2 = 32 − 2.32 y = 3 cos (50°) = 9 − 5.29 = 1.9 km

= 3.71y = 3.71

= 1.9 km

3 Write the answer in words. The boy walked a total of 1.9 km south of the starting point.

c 1 Draw a diagram of the journey and write in the distances found in parts a and b . The bearing of the boy from the starting point is represented by the angle α (that is, the angle measured in a clockwise direction from north to the line joining the starting and the fi nishing points of the journey).

c

2 km

3 km

N

130

N

1.9 km

4.3 kmO

A

2 The size of angle α cannot be found directly. Find the size of the supplementary angle labelled θ.

3 We have the lengths of the opposite side and the adjacent side, so choose the tangent ratio.

tan (θ)=OA

4 Substitute O = 4.3 and A = 1.9 and evaluate.

tan (θ) = 4.31.9

= 2.263 157 895

5 Make θ the subject of the equation using the inverse tangent function.

θ = tan−1 (2.263 157 895)

6 Evaluate and round to the nearest minute. = 66.161 259 82°= 66°9′40.535″= 66°10′

7 Find the angle α. α = 180° −66°10′= 113°50′

8 Write the answer in words. The bearing of the boy from his starting point is 113°50′ T.

180

To identify the compass bearing of an object we need to state (in this order):1. (a) whether the angle is measured from north (N) or south (S)(b) the size of the angle(c) whether the angle is measured in the direction of west (W) or east (E).True bearings are measured from north in a clockwise direction and expressed as 2. 3 digits.When solving problems involving bearings, always draw a clear diagram prior to 3. attempting the problem.

rememBer

Bearings 1 Change each of the following compass bearings to true bearings.

a N20°E b N20°W c S35°Wd S28°E e N34°E f S42°W

2 Change each of the following true bearings to compass bearings.

a 049°T b 132°T c 267°Td 330°T e 086°T f 234°T

3 Describe the following paths using true bearings.

a

35

N

3 km

b

2.5 km

N

S

W E22

c

8 km

N

S

W E35

d N

35 2.5 km4 km

e

7 km

12 km6550

NN

f

300 m

500 m

40 50

NN

exerCise

4G



4 Show each of the following by drawing the paths.

a A ship travels 040°T for 40 km and then 100°T for 30 km. b A plane flies for 230 km in a direction 135°T and a further 140 km in a direction 240°T.c A bushwalker travels in a direction 260°T for 0.8 km, then changes direction to 120°T for

1.3 km, and finally travels in a direction of 32° for 2.1 km.d A boat travels N40°W for 8 km, then changes direction to S30°W for 5 km and then

S50°E for 7 km.e A plane travels N20°E for 320 km, N70°E for 180 km and S30°E for 220 km.

5 We 17 a You are planning a trip on your yacht. If you travel 20 km from A to B on a bearing of 42°T: i how far east of A is B? ii how far north of A is B? iii what is the bearing of A from B?b In the next part of the journey you decide to travel 80 km from B to C on a bearing of

130°T. i Show the journey to be travelled using a diagram. ii How far south of B is C? iii How far east of B is C? iv What is the bearing of B from C?c In the next part of the journey you decide to travel 45 km from C to D on a bearing of

210°T. i Show the journey to be travelled using a diagram. ii How far south of C is D? iii How far west of C is D? iv What is the bearing of C from D?

6 If a farmhouse is situated 220 m N35°E from a shed, what is the true bearing of the shed from the house?

7 A pair of hikers travel 0.7 km on a true bearing of 240° and then 1.3 km on a true bearing of 300°. How far west have they travelled from their starting point?

8 We 18 A boat travels 6 km on a true bearing of 120° and then 4 km on a true bearing of 080°. a How far east is the boat from the starting point on the completion of its journey?b How far south is the boat from the starting point on the completion of its journey?c What is the bearing of the boat from the starting point on the completion of its journey?

9 A plane flies on a true bearing of 320° for 450 km. It then flies on a true bearing of 350° for 130 km and finally on a true bearing of 050° for 330 km. How far north of its starting point is the plane?

10 Find the final bearing for each of the following. Express your answer in true bearings, correct to the nearest minute.a A boat travels due east for 4 km

and then travels N20°E for 3 km. What is the final bearing of the boat from the starting point?

b A bushwalker travels due north for 3 km, then due east for 8 km. What is the final bearing of the bushwalker from the starting point?

c A car travels due south for 80 km, then travels due west for 50 km, and finally due south for a further 30 km. What is the final bearing of the car from the starting point?

182

The unit circle — quadrant 1A unit circle is a circle with centre at the origin and a radius of 1 unit.

–1 1

1

–1

y

x

1 unit

A right-angled triangle containing angle θ can be drawn in quadrant 1 of the unit circle as shown in the diagram below. (Note that angle θ is measured from the positive direction of the x-axis in an anticlockwise direction.)

–1 1

1

–1

y

xA

H

O

In this triangle, the hypotenuse is represented by the radius of the circle and, hence, the length of the hypotenuse (H) is 1 unit. The lengths of the opposite and adjacent sides can be found using trigonometric ratios as follows:

sin (θ) = OH

and cos (θ) = AH

sin (θ) = O (as H = 1) cos (θ)= A (as H = 1)

y

x

a sin ( )

b cos ( )

c 1

Thus the length of the side opposite to angle θ is sin (θ) units and the length of the side adjacent to angle θ is cos (θ) units.

As we have a unit circle, the hypotenuse is always 1 unit, but the lengths of the other two sides change, depending on the size of the angle θ.

4h



The pythagorean identityAs the triangle is a right-angled triangle, Pythagoras’ theorem c2 = a2 + b2 applies. Since a = sin (θ), b = cos (θ) and c = 1, substituting into Pythagoras’ theorem gives:

(sin (θ))2 + (cos (θ))2 = 12 This is known as the Pythagorean identity and is usually written as:

sin2 (θ) + cos2 (θ) = 1.Rearranging sin2 (θ) + cos2 (θ) = 1 gives other useful expressions.

cos2 (θ) = 1 − sin2 (θ) (and hence cos (θ) = 1 sin1 sin1 s ( )21 s−1 s ( )θ( ) in quadrant 1) and

sin2 (θ) = 1 − cos2 (θ) (and hence sin (θ) = 1 c ( )2− θ1 c− θ1 cos− θos ( )− θ( )2− θ2 in quadrant 1).

These results allow us to determine the value of either the sine ratio or the cosine ratio when we are given the other.

If cos (θ) = 0.3760, fi nd sin (θ). Give the answer correct to 4 decimal places.

Think WriTe

1 Write the transposed Pythagorean identity where sin (θ) is the subject.

sin (θ) = 1 2− cos (2s (2 )θ)θ)

2 Substitute cos (θ) = 0.3760. sin (θ) = 1 0 3760 21 0−1 0( .1 0( .1 0 )

3 Evaluate and round the answer, correct to 4 decimal places.

= 1 0 1413761 0−1 0.

= 0 858624. = 0.9266

If a tangent to the unit circle is drawn so that it is parallel to the y-axis, and the radius is extended until it meets with the tangent, then the length of the side opposite angle θ in the triangle formed is equal to tan (θ). The location of tan (θ) can be seen from the diagram at right.

By comparing similar triangles, we can determine the ratio for tan (θ).

sin( )

cos( )

tan( )

1is similar to

In similar triangles the corresponding sides are in the same ratio. Therefore,

sin ( )cos ( )

tan ( )θ)θ)θ)θ)

θ)θ)

θ θθ

=

=

1or

tan ( )sin ( )cos ( )

Worked example 19

y

x

1

tan( )si

n(

)

cos( )

184

If sin (θ) = 0.9396 and cos (θ) = 0.3420, fi nd tan (θ), correct to 4 decimal places.

Think WriTe

1 Write the identity that connects sin (θ), cos (θ) and tan (θ). tan (θ) =sin ( )cos ( )

θ)θ)θ)θ)

2 Substitute given values of sin (θ) and cos (θ) into the formula.

tan (θ) =0 93960 3420

.

.

3 Evaluate and round correct to 4 decimal places. = 2.7474

If sin (θ) = 0.5456: a fi nd the value of cos (θ), correct to 4 decimal placesb fi nd the value of tan (θ), correct to 4 decimal placesc fi nd the size of angle θ, correct to the nearest minuted draw a diagram to show angle θ, its sine, cosine and tangent.

Think WriTe/draW

a 1 To fi nd cos (θ), use the Pythagorean identity, transposed so that cos (θ) is the subject.

a cos (θ) = 1 2− sin (2n (2 )θ)θ)

2 Substitute sin (θ) = 0.5456. cos (θ) = 1 0 5456 21 0−1 0( .1 0( .1 0 )

3 Evaluate and round the answer, correct to 4 decimal places.

= 0 702 320 64.

= 0.8380

b 1 Write the identity for tan (θ). b tan (θ) =sin ( )cos ( )

θ)θ)θ)θ)

2 Substitute the values of sin (θ) and cos (θ) into the rule. =0 54560 8380

.

.

3 Evaluate and round the answer, to 4 decimal places. = 0.6511

c 1 As we are given the value of sin (θ), to fi nd the size of angle θ we must fi nd the inverse of sin (θ).

c θ = sin−1 (0.5456)

= 33.065 675 07°2 Convert to degrees and minutes by using a

calculator.= 33°3′56.43″= 33°4′

d Draw the unit circle. In quadrant 1 construct a right-angled triangle. Write in the size of the angle and show the location of the angle’s sine, cosine and tangent.

d y

x

1 unit

1 unit

sin(

33 4

')

tan(33 4')

cos(33 4')

33 4'

Worked example 20

Worked example 21



A unit circle has a radius of 1 unit.1. When a right-angled triangle is drawn in quadrant 1 of the unit circle, the length of the 2. side adjacent to the angle θ is equal to cos (θ) and the length of the side opposite to the angle θ is equal to sin (θ).

3. If a tangent to the unit circle is drawn so that it is parallel to the y-axis, and the radius is extended until it meets the tangent then, in the triangle formed, the length of the side opposite angle θ is equal to tan (θ).The Pythagorean identity is given by: 4. sin2 (θ) + cos2 (θ) = 1.The Pythagorean identity can be transposed to give the 5. following formulas for the fi rst quadrant:

sin (θ) = 1 2− cos (2s (2 )θ)θ) and cos (θ) = 1 2− sin (2n (2 )θ)θ)

The identity connecting the tangent of angle 6. θ with its sine and cosine is:

tan (θ) = sin ( )cos ( )

θ)θ)θ)θ)

.

y

xcos( )

sin

( )

tan

( )

1

–1

–1 1

rememBer

The unit circle — quadrant 1 1 We19 Find sin (θ) (correct to 4 decimal places) if:

a cos (θ) = 0.76 b cos (θ) = 0.87 c cos (θ) = 0.92d cos (θ) = 2

3 e cos (θ) = 13 f cos (θ) = 2

5 .

2 Find cos (θ) (correct to 4 decimal places) if:a sin (θ) = 0.852 b sin (θ) = 0.153 c sin (θ) = 1

2

d sin (θ) = 25 e 2 sin (θ) = 1.42 f 3 sin (θ) = 0.983.

3 We20 Find tan (θ) (correct to 3 decimal places) for each of the following.a sin (θ) = 0.5, cos (θ) = 0.866 b sin (θ) = 0.8, cos (θ) = 0.6c sin (θ) = 2, cos (θ) = 1

2 d sin (θ) = 15, cos (θ) = 0.8944

e sin (θ) = 0.8132, cos (θ) = 0.582 f sin (θ) = 0.9325, cos (θ) = 0.3612

4 Given that:a sin (θ) = 0.215, fi nd i cos (θ) and ii tan (θ) b cos (θ) = 0.992, fi nd i sin (θ) and ii tan (θ)c cos (θ) = 0.315, fi nd i sin (θ) and ii tan (θ) d sin (θ) = 0.876, fi nd i cos (θ) and ii tan (θ)e 5 cos (θ) = 4, fi nd i sin (θ) and ii tan (θ) f 7 sin (θ) = 3, fi nd i cos (θ) and ii tan (θ).

5 We21 If sin (θ) = 0.6112:a fi nd the value of cos (θ), correct to 4 decimal placesb fi nd the value of tan (θ), correct to 4 decimal placesc fi nd the size of angle θ, correct to the nearest minuted draw a diagram to show angle θ, its sine, cosine and tangent.

6 If cos (θ) = 0.215:a fi nd the value of sin (θ), correct to 4 decimal placesb fi nd the value of tan (θ), correct to 4 decimal placesc fi nd the size of angle θ, correct to the nearest minuted draw a diagram to show angle θ, its sine, cosine and tangent.

exerCise

4h

186

Circular functionsThe unit circle can be divided into 4 quadrants.

As you can see from the diagram, all angles in quadrant 1 are between 0° and 90°. All angles in quadrant 2 are between 90° and 180°, in quadrant 3 between 180° and 270°, and in quadrant 4 between 270° and 360°.

State the quadrant of the unit circle in which each of the following angles is found.a 145° b 282°

Think WriTe

a The given angle is between 90° and 180°. State the appropriate quadrant.

a 145° is in quadrant 2.

b The given angle is between 270° and 360°. State the appropriate quadrant.

b 282° is in quadrant 4.

So far we have looked at triangles constructed in quadrant 1 of the unit circle, with the angle θ being less than 90°. However, triangles can be drawn in other parts of the circle and we need to know what happens when angles become greater than 90°.

We can certainly use a calculator to fi nd sine, cosine and tangent values for angles greater than 90° , but it is important to understand where these values have come from.

In this section, we will look at fi nding sine and cosine values for angles greater than 90°, using a unit circle. As the unit circle is used to fi nd these values, sin (θ) and cos (θ) are often referred to as circular functions.

In the previous section we discovered that if a right-angledtriangle containing angle θ is constructed in quadrant 1 of the unit circle, then the value of sin (θ) can be found by measuring the length of the opposite side and the value of cos (θ) by measuring the length of the adjacent side.

The point of intersection of the radius (which is one of the arms of angle θ) with the unit circle, is P. From the diagram at right observe that cos (θ) represents the x-coordinate of point P and sin (θ) represents its y-coordinate. This observation provides us with the technique for fi nding sine and cosine of any angle in the unit circle, as shown below.

To fi nd the value of sine and/or cosine of any angle θ from the unit circle, follow these steps:1. Draw a unit circle.2. Construct the required angle so that its vertex is at the origin and the angle itself is

measured from 0° (as marked on the x-axis) in an anticlockwise direction. Label the point of intersection of the radius and the unit circle, P.

3. Use a ruler to fi nd the coordinates of point P.4. Interpret the results: x = cos (θ) and y = sin (θ), where x and y are coordinates of P.

4i y

x

1stquadrant

2ndquadrant

90

180360

0

270

3rdquadrant

4thquadrant

Worked example 22

eBookpluseBookplus

Interactivityint-1414

Circular functions

y

x

90

1

1

270

1801

0

3601cos( )

sin( )

P



Find the value of each of the following using the unit circle.a sin (200°) b cos (200°)

Think WriTe/draW

Draw a unit circle and construct an angle of 200°. Label the point corresponding to the angle of 200° on the circle P. Highlight the lengths, representing the x- and y-coordinates of point P.

P

y

x

90

1

–1

270

180–1

0

3601

= 200

yx

a The sine of the angle is given by the y-coordinate of P. Find the y-coordinate of P by measuring the distance along the y-axis. State the value of sin (200°). (Note that the sine value will be negative as the y-coordinate is negative.

a sin (200°) = −0.3

b The cosine of the angle is given by the x-coordinate of P. Find the x-coordinate of P by measuring the distance along the x-axis. State the value of cos (200°). (Note that cosine is also negative in quadrant 3, as the x-coordinate is negative.)

b cos (200°) = −0.9

The results obtained in Worked example 23 can be verified with the aid of a calculator: sin (200°) = −0.342 020 143 and cos (200°) = −0.939 692 62.

Rounding these values to 1 decimal place would give −0.3 and −0.9 respectively, which match the values obtained from the unit circle.

angles in degrees and radiansSo far, the unit we have used to measure angles is the degree (°). Another angle unit is the radian (c). You will recall that, in using a CAS calculator to determine trigonometric ratios, you were advised to ensure that your calculator was in Degree mode. Another option is the radian mode.

Consider a unit circle — one with a radius of 1 unit.

Let the radius OP rotate anticlockwise to a point P′ on the circumference of the circle where the arc length PP′ is 1 unit in length (the same as the length of the radius).

Worked example 23

O P

OP = 1 unit

y

x

O P

P

c

OP = 1 unit

y

x

188

The length of the arc PP′ represents a measurement of 1 radian. Since this is a length on the circumference of a circle, we can fi nd a relationship between a radian and π. Consider the radius rotated 180° around the circumference of the circle.

Length of semicircular arc PP′ = 2

2πr

= 2 1

22 1× ×2 1π2 1π2 12 1× ×2 1π2 1× ×2 1

= πc

From this we can see that: 180° = πc

1° = π c

180

To convert an angle in degrees to radian measure, multiply by π c

180°.

Also, since 180° = πc, it follows that 1c = 180°

π.

To convert an angle in radian measure to degrees, multiply by 180

c

°π

.

Find the radian measure that corresponds to the following turns around a unit circle.

a 12 turn b

13 turn

Think WriTe

a 1 Find the number of degrees in this turn. a 12 turn = 1

2 × 360°

= 180°

2 To convert an angle in degrees to radian

measure, multiply the angle by π c

180°.

180° = π

180 × 180

= πc

3 Write the answer.12 turn is equal to πc.

b 1 Find the number of degrees in this turn. b 13 turn =

13 × 360°

= 120°

2 To convert an angle in degrees to radian


180°.

120° = π

180 × 120

= 120180

π

=23π c

3 Write the answer.13 turn is equal to

23π c

.

O PP

1—2

circumference

180

y

x

To convert an angle in degrees to radian measure, multiply by

eBookpluseBookplus


Degrees and radians

Worked example 24



a Convert 150° to radian measure, expressing the answer in terms of π.

b Convert the radian measurement 3

4πππππ c

to degrees.

Think WriTe

a 1 To convert an angle in degrees to radian


180°.

a 150° = π

180 × 150

2 Simplify, leaving the answer in terms of π. = 5

6π c

b 1 To convert an angle in radian measure to

degrees, multiply the angle by 180°π c .

b 34π c

=180π

× 34π

2 Simplify and write the answer.Note: The π cancels out.

= 135°

The unit circle is divided into four quadrants, as shown.1.

y

x

2ndquadrant

1stquadrant

90

180360

0

270

3rdquadrant

4thquadrant

y

x

90

1

1

270

1801

0

3601cos( )

sin( )

P

Sine and cosine of any angle, 2. θ, are given as follows: x = cos (θ) and y = sin (θ), where x and y are coordinates of point P on the unit circle, corresponding to the given angle. −3. 1 ≤ sin (θ) ≤ 1 and −1 ≤ cos (θ) ≤ 1.Sine is positive in quadrants 1 and 2 and negative in quadrants 3 and 4.4. Cosine is positive in quadrants 1 and 4 and negative in quadrants 2 and 3.5. Tangent is positive in quadrants 1 and 3 and negative in quadrants 2 and 4.6. An angle can be measured in degrees or radians.7.

To convert from degrees to radians, multiply by 8. π c

180°.

To convert from radians to degrees, multiply by 9. 180°π c .

rememBer

Worked example 25

190

Circular functions 1 We22 State which quadrant of the unit circle each of the following angles is in.

a 60° b 130° c 310° d 260°e 100° f 185° g 275° h 295°

2 We23 Draw a unit circle using a protractor and graph paper and use it to determine the value of each of the following.

a sin (20°) b cos (20°) c cos (100°) d sin (100°)e sin (320°) f cos (320°) g sin (215°) h cos (215°)

3 Use the unit circle to find each of the following.

a sin (90°) b cos (90°) c sin (180°) d cos (180°)e sin (270°) f cos (270°) g sin (360°) h cos (360°)

4 On the unit circle, use a protractor to measure an angle of 30° from the positive x-axis. Mark the point P on the circle. Use this point to construct a triangle in quadrant 1 as shown.a Find cos (30°). (Remember that the length of the adjacent

side of the triangle is cos (30°).)b Find sin (30°). (This is the length of the opposite side of

the triangle.)c Check your answers to a and b by finding these values

with a calculator.

5 Using your graph of the unit circle, measure 150° with a protractor and mark the point P on the circle. Use this point to draw a triangle in quadrant 2 as shown.a What angle does the radius OP make with the negative

x-axis?b Remembering that x = cos (θ), use your circle to find

the value of cos (150°).c How does cos (150°) compare to cos (30°)?d Remembering that y = sin (θ), use your circle to find

the value of sin (150°).e How does sin (150°) compare with sin (30°)?

6 On the unit circle, measure 210° with a protractor and mark the point P on the circle. Use this point to draw a triangle in quadrant 3 as shown.a What angle does the radius OP make with the negative

x-axis?b Use your circle to find the value of cos (210°).c How does cos (210°) compare to cos (30°)?d Use your circle to find the value of sin (210°).e How does sin (210°) compare with sin (30°)?

7 On the unit circle, measure 330° with a protractor and mark the point P on the circle. Use this point to draw a triangle in quadrant 4 as shown.a What angle does the radius OP make with the positive

x-axis?b Use your circle to find the value of cos (330°).c How does cos (330°) compare to cos (30°)?d Use your circle to find the value of sin (330°).e How does sin (330°) compare with sin (30°)?

exerCise

4i

y

xcos(30 )

sin(30 )30

P

O

y

xcos(150 )

sin(150 )150

P

O

y

x

cos(210 )

sin(210 )

210

P

O

y

x

cos(330 )

sin(330 )

330

P

O



8 On the unit circle, draw an appropriate triangle for the angle of 20° in quadrant 1.a Find sin (20°).b Find cos (20°).c Draw a tangent line and extend the hypotenuse of the

triangle to meet the tangent as shown. Accurately measure the length of the tangent between the x-axis and the point where it meets the hypotenuse and, hence, state the value of tan (20°).

d What is the value of sin ( )cos ( )

2020

°°

?

e How does tan (20°) compare with sin ( )cos ( )

2020

°°

?

9 On the unit circle, draw an appropriate triangle for the angle of 135° in quadrant 2.a Find sin (135°).b Find cos (135°).c Draw a tangent line and extend the hypotenuse of the

triangle to meet the tangent as shown. Accurately measure the length of the tangent to where it meets the hypotenuse to fi nd the value of tan (135°).


135135

°°

?


135135

°°

?

f How does tan (135°) compare with tan (45°)? 10 On the unit circle, draw an appropriate triangle for the angle of 220°

in quadrant 3.a Determine sin (220°).b Determine cos (220°).c Draw a tangent line and extend the hypotenuse of the

triangle to meet the tangent as shown. Calculate tan (220°) by accurately measuring the length of the tangent to where it meets the hypotenuse.


220220

°°

?


220220

°°

?

f How does tan (220°) compare with tan (40°)? (Use a calculator.)

11 On the unit circle, draw an appropriate triangle for the angle of 300° in quadrant 4.a Determine sin (300°).b Determine cos (300°).c Draw a tangent line and extend the hypotenuse of the

triangle to meet the tangent as shown. Calculate tan (300°) by accurately measuring the length of the tangent to where it meets the hypotenuse.

y

xcos(20 )

sin

(20

)

tan

(20

)

20

y

x

tan

(135

)

135

y

x

tan

(220

)

220

y

x

tan

(300

)

300

192


300300

°°

?


300300

°°

?

f How does tan (300°) compare with tan (60°)? (Use a calculator.)

12 We24 Find the radian measures that correspond to the following turns around a unit circle.

a 14 turn b 1

6 turn c 23 turn

13 We25 Convert the following angles to radian measure, expressing answers in terms of π.

a 30° b 72° c 225°d 200° e 90° f 48°

14 Convert the following radian measures into degrees.

a π4

c

b π3

c

c 32π c

d 23π c

e π5

c

f 74π c

Graphs of trigonometric functionsGraphs of the circular functions y = sin (x) and y = cos (x) and others will be studied in detail in Years 11 and 12. This year an exploratory exercise has been included so that you can investigate the shape and some of the main features of the graphs of these trigonometric functions. The exercise has been designed as a sequence of related questions, leading you step by step in your exploration of the graphs of y = sin (x) and y = cos (x).

Graphs of trigonometric functions 1 Using your calculator (or the unit circle if you prefer), complete the following table.

x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°sin (x)

x 390° 420° 450° 480° 510° 540° 570° 600° 630° 660° 690° 720°sin (x)

2 On graph paper, rule x- and y-axes and carefully mark a scale along each axis. Use 1 cm = 30° on the x-axis to show x-values from 0° to 720°. Use 2 cm = 1 unit along the y-axis to show y-values from −1 to 1. Carefully plot the graph of y = sin (x) using the values from the table in question 1.

3 How long does it take for the graph of y = sin (x) to complete one full cycle?

4 From your graph of y = sin (x), find the value of y for each of the following.

a x = 42° b x = 130° c x = 160°d x = 200° e x = 180° f x = 70°g x = 350° h x = 290°

5 From your graph of y = sin (x), find the value of x for each of the following.

a y = 0.9 b y = −0.9 c y = 0.7d y = −0.5 e y = −0.8 f y = 0.4

eBookpluseBookplus


Circular functions

4J

exerCise

4J



6 Using your calculator (or the unit circle if you prefer), complete the following table.

x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°cos (x)

x 390° 420° 450° 480° 510° 540° 570° 600° 630° 660° 690° 720°cos (x)

7 On graph paper, rule x- and y-axes and carefully mark a scale along each axis. Use 1 cm = 30° on the x-axis to show x-values from 0° to 720°. Use 2 cm = 1 unit along the y-axis to show y-values from −1 to 1. Carefully plot the graph of y = cos (x) using the values from the table in question 6.

8 If you were to continue the graph of y = cos (x), what shape would you expect it to take?

9 Is the graph of y = cos (x) the same as the graph of y = sin (x)? How does it differ? What features are the same?

10 Using the graph of y = cos (x), find a value of y for each of the following.a 48° b 170° c 180° d 340°e 240° f 140° g 40° h 165°

11 Using the graph of y = cos (x), find a value of x for each of the following.a y = −0.5 b y = 0.8 c y = 0.7d y = −0.6 e y = 0.9 f y = −0.9

applicationsMany people use trigonometry at work. It is particularly important in careers such as the building trades, surveying, architecture and engineering. As you saw in section 4F, trigonometric ratios have a variety of applications. Not only can they be used to fi nd angles of elevation and depression, but also to calculate distances which we could not otherwise easily measure. When solving a problem, remember the following steps (mentioned in section 4F). 1. Sketch a diagram to represent

the situation described in the problem.

2. Label the sides of the right-angled triangle with respect to the angle involved.

3. Identify what is given and what needs to be found.

4. Select an appropriate trigonometric ratio and use it to fi nd the unknown measurement.

5. Interpret your result by writing a worded answer.

4k

194

A ladder of length 3 m makes an angle of 32° with the wall.a How far is the foot of the ladder from the wall?b How far up the wall does the ladder reach?c What angle does the ladder make with the ground?

Think WriTe/draW

Sketch a diagram and label the sides of the right-angled triangle with respect to the given angle.

3 m

(wall)

32 A

H

Ox

y

a 1 We need to fi nd the distance of the foot of the ladder from the wall (O) and are given the length of the ladder (H). Choose the sine ratio.

a sin (θ)=OH

2 Substitute O = x, H = 3 and θ = 32°. sin (32°) =x3

x = 3 sin (32°) ≈ 1.59 m

3 Make x the subject of the equation.

4 Evaluate and round the answer to 2 decimal places.

5 Write the answer in words. The foot of the ladder is 1.59 m from the wall.

b 1 We need to fi nd the height the ladder reaches up the wall (A) and are given the hypotenuse (H). Choose the cosine ratio.

b cos (θ)=AH

2 Substitute A = y, H = 3 and θ = 32°. cos (32°) =y3

3 Make y the subject of the equation. y = 3 cos (32°)

4 Evaluate and round the answer to 2 decimal places.

y ≈ 2.54 m

5 Write the answer in words. The ladder reaches 2.54 m up the wall.

c 1 To fi nd the angle that the ladder makes with the ground, we could use any of the trigonometric ratios, as the lengths of all three sides are known. However, it is quicker to use the angle sum of a triangle.

c α + 90° + 32° = 180°α + 122° = 180°

α = 180° − 122°α = 58°

2 Write the answer in words. The ladder makes a 58° angle with the ground.

Worked example 26



To solve a problem involving trigonometric ratios, follow these steps:Draw a diagram to represent the situation.1. Label the diagram with respect to the angle involved (either given or that needs to be 2. found).Identify what is given and what needs to be found.3. Select an appropriate trigonometric ratio and use it to find the unknown side or angle.4. Interpret the result by writing a worded answer.5.

rememBer

applications 1 We26 A 3 m-long ladder is placed against a wall so that it reaches 1.8 m up the wall.

a What angle does the ladder make with the ground?b What angle does the ladder make with the wall?c How far from the wall is the foot of the ladder?

2 Jamie decides to build a wooden pencil box. He wants his ruler to be able to lie across the bottom of the box, so he allows 32 cm along the diagonal. The width of the box is to be 8 cm.

8 cm32 cm

Calculate: a the size of angle θ b the length of the box.

3 A chord of a circle subtends an angle 80°56′ at the centre. If the chord is 31 cm long, how far is it from the centre?

31 cm

80 56´

4 A chord, AC, of a circle is inclined to the diameter, AB, at an angle of 24°35′.

26 cm

A C

B

24 35'

a If BC = 26 cm, calculate the length of the chord, AC.b Calculate the diameter of the circle.

exerCise

4k

196

5 A carpenter wants to make a roof pitched at 29°30′, as shown in the diagram. How long should he cut the beam, PR?

10.6 m

29 30'P Q

R

6 The sloping sides of a gable roof are each 7.2 m long. They rise to a height of 2.4 m in the centre. What angle do the sloping sides make with the horizontal?

7 The mast of a boat is 7.7 m high. A guy wire from the top of the mast is fixed to the deck 4 m from the base of the mast. Determine the angle the wire makes with the horizontal.

8 A desk top of length 1.2 m and width 0.5 m rises to 10 cm.

E F

DC

BA

0.5 m

1.2 m

10 cm

Calculate:a ∠DBFb ∠CBE.

9 A cuboid has a square end.

45 cm

H G

F

BA

D CE

O

25 cm

X

a If the length of the cuboid is 45 cm and its height and width are 25 cm each, calculate: i the length of BD ii the length of BG iii the length of BE iv the length of BH v ∠FBG vi ∠EBH.b If the midpoint of FG is X and the centre of the square ABFE is O calculate: i the length OF ii the length FX iii ∠FOX iv the length OX.

10 In a right square-based pyramid, the length of the side of the base is 12 cm and the height is 26 cm.

12 cm

26 c

m



Determine:a the angle the triangular face makes with the baseb the angle the sloping edge makes with the basec the length of the sloping edge.

11 In a right square-based pyramid, the length of the side of the square base is 5.7 cm.

5.7 cm

68

If the angle between the triangular face and the base is 68°, determine:a the height of the pyramidb the angle the sloping edge makes with the basec the length of the sloping edge.

12 In a right square-based pyramid, the height is 47 cm. If the angle between a triangular face and the base is 73°, calculate:a the length of the side of the square baseb the length of the diagonal of the basec the angle the sloping edge makes with the base.

13 The height of a vertical cone is 24.5 cm.

24.5 cm

48 37'10"

If the angle at the apex is 48°37′10″, determine:a the length of the slant edge of the coneb the radius of the cone.

198

sUmmary

Pythagoras’ theorem

The hypotenuse is the longest side of the triangle and is opposite the right angle. 1. On your diagram, check whether you are fi nding the length of the hypotenuse or one of the shorter sides. 2. The length of the hypotenuse can be found if we are given the length of the two shorter sides by using the 3. formula c2 = a2 + b2.The length of the shorter side can be found if we are given the length of the hypotenuse and the other 4. shorter side by using the formula: a2 = c2 − b2 or b2 = c2 − a2. When using Pythagoras’ theorem, always check the units given for each measurement. 5. If necessary, convert all measurements to the same units before using the rule.6. Worded problems can be solved by drawing a diagram and using Pythagoras’ theorem to solve the problem. 7. Worded problems should be answered in a sentence. 8.

Pythagoras’ theorem in three dimensions

Pythagoras’ theorem can be used to solve problems in three dimensions (3-D).1. Some common 3-D shapes include boxes, pyramids and right-angled wedges.2. To solve problems in 3-D it is helpful to draw sections of the original shape in two dimensions (2-D).3.

Trigonometric ratios

When using the calculator to fi nd values of sine, cosine and tangent, make sure the calculator is in Degree 1. mode.To fi nd the size of an angle whose sine, cosine or tangent is given, perform an inverse operation; that is, 2. sin

−1, cos−1 or tan

−1.Use the calculator’s conversion function to convert between decimal degrees and degrees, minutes and 3. seconds.There are 60 minutes in 1 degree and 60 seconds in 1 minute.4. The three trigonometric ratios, sine, cosine and tangent, are defi ned as:5.

sin (θ)= OH

, cos (θ)=AH

and tan (θ)= OA

,

where H is the hypotenuse, O is the opposite side and A is the adjacent side.The three ratios are abbreviated to the useful mnemonic SOH CAH TOA.6. To determine which trigonometric ratio to use, follow these steps.7. (a) Label the sides of the right-angled triangle that are either given, or need to be found, using the

symbols O, A, H with respect to the angle in question.(b) Consider the sides that are involved and write the trigonometric ratio containing both of these sides.

(Use SOH CAH TOA to assist you.)(c) Identify the values of the pronumerals in the ratio.(d) Substitute the given values into the ratio.

Using trigonometry to calculate side lengths

The trigonometric ratios can be used to fi nd a side length in a right-angled triangle when we are given 1. other side length and one of the acute angles.The calculator step will differ depending upon whether the unknown is in the numerator or denominator 2. of the equation formed after substitution.

Using trigonometry to calculate angle size

The trigonometric ratios can be used to fi nd the size of the acute angles in a right-angled triangle when we 1. are given the length of two sides.To fi nd an angle size we need to use the inverse trigonometric functions.2. Answers may be given correct to the nearest degree, minute or second.3.



Angles of elevation and depression

The angle of elevation is measured up and the angle of depression is measured down from the horizontal 1. line to the line of vision.

Angle ofelevation

Horizontal

HorizontalAngle ofdepression

For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of 2. depression of A as seen from B.

A



B

Bearings

To identify the compass bearing of an object we need to state (in this order):1. (a) whether the angle is measured from north (N) or south (S)(b) the size of the angle(c) whether the angle is measured in the direction of west (W) or east (E).True bearings are measured from north in a clockwise direction and expressed as 3 digits.2. When solving problems involving bearings, always draw a clear diagram prior to attempting the problem.3.

The unit circle — quadrant 1

A unit circle has a radius of 1 unit.1. When a right-angled triangle is drawn in quadrant 1 of the unit circle, the length of the side adjacent to the 2. angle θ is equal to cos (θ) and the length of the side opposite to the angle θ is equal to sin (θ). If a tangent to the unit circle is drawn so that it is parallel to the 3. y-axis, and the radius is extended until it meets the tangent then, in the triangle formed, the length of the side opposite angle θ is equal to tan (θ).

y

xcos( )

sin

( )

tan

( )

1

–1

–1 1

The Pythagorean identity is given by: sin4. 2 (θ) + cos2 (θ) = 1.The Pythagorean identity can be transposed to give the following formulas for the fi rst quadrant:5.

sin (θ) = 1 2− cos (2s (2 )θ)θ) and cos (θ) = 1 2− sin (2n (2 )θ)θ).

The identity connecting the tangent of angle 6. θ with its sine and cosine is:

tan (θ) = sin ( )cos ( )

θ)θ)θ)θ)

.

200

Circular functions

The unit circle is divided into four quadrants, as shown.1.

y

x

2ndquadrant

1stquadrant

90

180360

0

270

3rdquadrant

4thquadrant

y

x

90

1

1

270

1801

0

3601cos( )

sin( )

P

Sine and cosine of any angle, 2. θ, are given as follows: x = cos (θ) and y = sin (θ), where x and y are coordinates of point P on the unit circle, corresponding to the given angle. −3. 1 ≤ sin (θ) ≤ 1 and −1 ≤ cos (θ) ≤ 1.Sine is positive in quadrants 1 and 2 and negative in quadrants 3 and 4.4. Cosine is positive in quadrants 1 and 4 and negative in quadrants 2 and 3.5. Tangent is positive in quadrants 1 and 3 and negative in quadrants 2 and 4.6. An angle can be measured in degrees or radians.7.

To convert from degrees to radians, multiply by 8. π c

180°.

To convert from radians to degrees, multiply by 9. 180°π c .

Applications

To solve a problem involving trigonometric ratios, follow these steps:Draw a diagram to represent the situation.1. Label the diagram with respect to the angle involved (either given or that needs to be found).2. Identify what is given and what needs to be found.3. Select an appropriate trigonometric ratio and use it to fi nd the unknown side or angle.4. Interpret the result by writing a worded answer.5.

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ChapTer reVieW

mUlTiple ChoiCe

1 The most accurate measure for the length of the third side in the triangle at right is:A 4.83 mB 23.3 cmC 3.94 mD 2330 mmE 4826 mm

2 What is the value of x in this figure?A 5.4B 7.5C 10.1D 10.3E 4

3 What is the closest length of AG of the cube at right?A 10B 30C 20D 14E 17

4 If sin (38°) = 0.6157, which of the following will also give this result?A sin (218°) B sin (322°)C sin (578°) D sin (682°)E sin (142°)

5 The angle 118°52′34″ is also equal to:

A 118.5234° B 1185234

°

C 118.861° D 118.876°E 118.786°

6 Which trigonometric ratio for the triangle shown at right is incorrect?

A sin (α) = bc B sin (α) =

ac

C cos (α) = acD tan (α) = b

aE tan (θ) = a

b 7 Which of the following statements is correct?

A sin (55°) = cos (55°)B sin (45°) = cos (35°)

C cos (15°) = sin (85°)D sin (30°) = sin (60°)E sin (42°) = cos (48°)

8 Which of the following can be used to find the value of x in the diagram below?

28.7

x35

A 28.7 sin (35°) B 28.7 cos (35°)

C 28.7 tan (35°) D 28 735.

sin ( )°E

28 735.

cos ( )°

9 Which of the following expressions can be used to find the value of a in the triangle shown?

a

75 35

A 35 sin (75°) B sin−1 35

75

C sin−1 75

35

D cos

−1 3575

E cos−1 75

35

10 If a school is 320 m S42°W from the police station, what is the true bearing of the police station from the school?A 042°TB 048°TC 222°TD 228°TE 312°T

11 If tan (θ) = 0.3652, then:A sin (θ) = 0.6348, cos (θ) = 0.3652B sin (θ) = 0.3652, cos (θ) = 0.6348C sin (θ) = 0.3652, cos (θ) = 0.3652D sin (θ) = 0.3430, cos (θ) = 0.9393E sin (θ) = 0.9393, cos (θ) = 0.3430

2840 mm

5.6 m

x5

2 7

A B

C

F

GH

D

E

10

10

10

ba

c

202

shorT ansWer

1 Calculate x, correct to 2 decimal places.a

123.1 cm

48.7 cm

x

b 117 mm

82 mm x

2 Calculate the value of the pronumeral, correct to 2 decimal places.

3 Calculate the height of this pyramid.

4 A person standing 23 m away from a tree observes the top of the tree at an angle of elevation of 35°. If the person is 1.5 m tall, what is the height of the tree?

5 A man of height 1.8 m stands at the window of a tall building. He observes his young daughter in the playground below. If the angle of depression from the man to the girl is 47° and the floor on which the man stands is 27 m above the ground, how far from the bottom of the building is the child?

6 A plane flies 780 km in a direction of 185°T. How far west has it travelled from the starting point?

7 A hiker travels 3.2 km on a bearing of 250°T and then 1.8 km on a bearing of 320°T. How far west has she travelled from the starting point?

8 a If sin (θ) = 0.423 and cos (θ) = 0.906, find tan (θ).

b If sin (θ) = 0.988 and cos (θ) = 0.070, fi nd tan (θ).

9 If a 4 m ladder is placed against a wall and the foot of the ladder is 2.6 m from the wall, what angle does the ladder make with the wall?

x x

13.4 cm

8 mm

10 mm

8 mm

exTended response

1 A surveyor needs to determine the height of a building. She measures the angle of elevation of the top of the building from two points, 64 m apart. The surveyor’s eye level is 195 cm above the ground.

x

h

36 24'47 48'195 cm64 m

a Find the expressions for the height of the building, h, in terms of x using the two angles.

b Solve for x by equating the two expressions obtained in part a.c Find the height of the building.

2 The height of a right square-based pyramid is 13 cm. If the angle the face makes with the base is 67°, find:a the length of the edge of the square baseb the length of the diagonal of the basec the angle the slanted edge makes with the base.

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Test YourselfChapters 1–4



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Are you ready?Digital docs (page 145)

SkillSHEET 4.1: Rounding to a given number of • decimal placesSkillSHEET 4.2: Rounding the size of an angle to • the nearest minute and secondSkillSHEET 4.3: Labelling the sides of a right-• angled triangleSkillSHEET 4.5: Rearranging formulas• SkillSHEET 4.6: Drawing a diagram from given • directions

4A Pythagoras’ theoremDigital doc

SkillSHEET 4.1: Rounding to a given number of • decimal places (page 149)

4B Pythagoras’ theorem in three dimensionsDigital docs

SkillSHEET 4.2: Drawing 3-D shapes • (page 155)WorkSHEET 4.1: Pythagoras’ theorem • (page 157)

4C Trigonometric ratiosDigital docs

SkillSHEET 4.3: Labelling the sides of a • right-angled triangle (page 162)SkillSHEET 4.4: Selecting an appropriate • trigonometric ratio based on the given information (page 163)

4D Using trigonometry to calculateside lengths

Interactivity int-1146Using trigonometry • (page 164)

4E Using trigonometry to calculate angle sizeDigital docs

SkillSHEET 4.7: Rounding angles to the nearest • degree (page 170)WorkSHEET 4.2: Using trigonometry• (page 171)

4F Angles of elevation and depressioneLesson eles-0173

Height of a satellite • (page 172)Digital docs

SkillSHEET 4.8: Drawing a diagram from given • directions (page 174)WorkSHEET 4.3: Elevation and depression • (page 176)

4I Circular functionsInteractivities

Circular functions • int-1414 (page 186)Degrees and radians • int-1413 (page 188)

Digital doc

WorkSHEET 4.4: Circular functions • (page 192)

Chapter summaryInteractivities (page 200)

Word search Chapter 4 • (int-1095): An interactive word search involving words associated with the chapter.Crossword Chapter 4 • (int-1108): An interactive crossword using the defi nitions associated with the chapter.

Chapter reviewInteractivities (page 202)

Test Yourself Chapter 4 • (int-1121): An interactive test covering the concepts from this chapter.Cumulative Test Yourself Chapters 1–4 • (int-1133): A cumulative interactive test covering content from this chapter and all previous chapters.

To access eBookPLUS activities, log on to

www.jacplus.com.au

ChapTer reVieW

204

WorkinG maThemaTiCally 1

TimeThe measuring of time has evolved from ancient methods, such as sundials, to the modern techniques of using the Global Positioning System in coordination with the Network Time Protocol to synchronise timekeeping systems across the globe.

In the early seventeenth century, Galileo Galilei was able to use the regular motion of a pendulum as a method of keeping time. This discovery opened the way for further mathematical discoveries based around the simple device of a pendulum.

Where to from here?kinematicsmechanics

1 How does a pendulum clock work?Historically, pendulum clocks were among the first mechanical timepieces. Explain how they work.

2 What is a simple pendulum?

3 What are the properties of a pendulum that we can investigate?(a) In making a pendulum and setting it in motion, what things can we change?(b) What can we measure?(c) What can we control?(d) Which variables affect the period?

4 How can we build a physical model of a simple pendulum?Design and build a simple pendulum that can be used to enable measurement of its period for a variety of variables, including lengths up to 2 m.

Test tube clamp

Retort standCotton

Bob

Period

Angulardisplacement

Vertical


Working mathematically 1 205

5 What is the best way to measure the period of a pendulum?Design a procedure to obtain as accurate a measure as possible of the period for a pendulum with a length of 1 m. It is absolutely imperative that you work to reduce operator error.

6 How do you design a pendulum with a period of 1 second?By taking no more than three sets of measurements relating a pendulum’s length to its period, determine the length for period of 1 s.

7 How does a computer model compare with a real pendulum?Conduct an investigation to assess how well a computer simulation matches the results for a physical pendulum by using the Pendulum interactivity in your eBookPLUS.

8 How does period vary with length?Design an experiment that will result in a graph of the relationship between the length of a simple pendulum and its period. You may use a real pendulum, a computer simulation or both. Justify your choice. Decide on:• the procedure you will use• the number of lengths to consider• the spacing of the measurements• how many readings you will take for any one length.

9 Which algebraic function is the best model for the relationship between pendulum length and period?(a) Draw a scatterplot of the data points you have collected relating pendulum

length to period. (b) How well is the data matched by linear, power and quadratic functions?

(c) It has been suggested that the mathematical model Tlg

= 2π can be used

to calculate the period of a simple pendulum. T is the period in seconds, l is the length in metres and g is the Earth’s gravitational fi eld strength. (The gravitational fi eld strength measures how strongly the Earth’s gravitational force attracts other masses. Its value is relatively constant at the Earth’s surface, where it is approximately 9.80 newtons per kilogram.) Investigate this model.

(d) One student has said that T l≈ 2 . This would mean that if l = 1 then t ≈ 2. Is that close? Why does that approximation work?

10 What does the complete graph relating pendulum length and period look like?

(a) By selecting convenient lengths between 10 cm and 2 m,

sketch a graph of Tlg

= 2π .

(b) Use this graph to explain why linear extrapolation often produces an inaccurate result.

11 How do pendulums work elsewhere in the solar system?How would the results differ on the moon?What if we could conduct the experiment on Mars?

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Pendulum

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Pendulum scatterplot

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