MFM2P - WordPress.comMFM2P – Foundations of Mathematics Unit 1 - Introduction Grade 10 Mathematics...

MFM2P

Foundations of Mathematics Grade 10 Applied

MFM2P – Foundations of Mathematics Unit 1 - Introduction

Grade 10 Mathematics (Applied)

Welcome to the Grade 10 Foundations of Mathematics, MFM 2P. This full-credit course is part of the new Ontario Secondary School curriculum. This course enables students to develop and understanding of the mathematical concepts related to introductory algebra, proportional reasoning, and measurement and geometry through investigation and the effective use of technology. Students will investigate real-life examples to develop various representations of linear relationships, and will determine the connections between the representations. They will also explore certain relationships that emerge from the measurement of three-dimensional figures and two-dimensional shapes. Students will consolidate their mathematical skills as they solve problems and communicate their thinking.

Materials

This course is self-contained and does not require a textbook. You will require lined paper, graph paper, a ruler, a scientific calculator and a writing utensil.

Expectations

The overall expectations you will cover in the lesson are listed on the first page of each lesson.

Lesson Description

Each lesson contains one or more concepts with each being followed by support questions. At the end of the lesson the key questions covering all concepts in the lesson are assigned and will be submitted for evaluation.

Evaluation

In each lesson, there are support questions and key questions. You will be evaluated on your answers to the key questions in each lesson, the mid-term exam and the final exam.

Support Questions

These questions will help you understand the ideas and master the skills in each lesson. They will also help you improve the way you communicate your ideas. The support questions will prepare you for the key questions.

Copyright © 2005, Durham Continuing Education of 57


Write your answers to the support questions in your notebook. Do not submit these answer for evaluation. You can check your answers against the suggested answers that are given at the end of each unit.

Key Questions The key questions evaluate your achievement of the expectations for the lesson. Your answers will show how well you have understood the ideas and mastered the skills. They will also show how well you communicate your ideas. You must try all the key questions and complete most of them successfully in order to pass each unit. Write your answers to the key questions on your own paper and submit them for evaluation at the end of each unit. Make sure each lesson number and question is clearly labeled on your submitted work. Mid-term and Final Examination In this course there is a mid-term and final exam. These exams will incorporate the four learning categories knowledge and understanding, application, communication, and thinking and inquiry.



Unit One Lesson One

Similar triangle notation Interior angle property of similar triangles Proportionate side property of similar triangle

Lesson Two

Triangle notation Substitution into the Pythagorean theorem Using Pythagorean theorem to find the missing side of a triangle

Lesson Three

Label parts of a right triangle i.e. opposite, adjacent and hypotenuse Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse Using the sine ratio to find unknown sides of a right triangle Using the sine ratio to find unknown interior angles of a right triangle

Lesson Four

Label parts of a right triangle i.e. opposite, adjacent and hypotenuse Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse Using the cosine ratio to find unknown sides of a right triangle Using the cosine ratio to find unknown interior angles of a right triangle

Lesson Five

Label parts of a right triangle i.e. opposite, adjacent and hypotenuse Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse Using the tangent ratio to find unknown sides of a right triangle Using the tangent ratio to find unknown interior angles of a right triangle



Unit Two Lesson Six

Explain and use correctly prefixes in the imperial and metric system Convert between imperial and metric units commonly used in everyday

applications

Lesson Seven

Radius and diameter Calculations using pi (π) Solving volume questions using formulas and substitution

Lesson Eight

Introduction to surface area Radius and diameter Calculations using pi (π) Solving surface area questions using formulas and substitution

Lesson Nine

Continued introduction to algebra Solving for unknowns Checking solutions to algebraic equations

Lesson Ten

Using algebra to convert to y-intercept form from standard form Using algebra to convert to standard form from y-intercept form



Unit Three Lesson Eleven

Introduction to the line Using standard form of an equation Using y-intercept form of an equation x and y intercept Recognizing positive, negative, zero and undefined slopes Using the rise and the run of a given line to find its slope Using a pair of coordinates of a line to calculate slope

Lesson Twelve

Converting equation of a line from y-intercept form to standard form Recognizing and graphing parallel slopes Recognizing and graphing perpendicular slopes

Lesson Thirteen

Graphing linear equations Checking whether a coordinate pair satisfies and equation Finding the coordinates of the point of intersection of two equations Recognizing the meaning of the point of intersection of a linear system of

equations

Lesson Fourteen

Isolating a variable Checking whether a coordinate pair satisfies and equation Finding the coordinates of the point of intersection of two equations

using algebra Recognizing the meaning of the point of intersection of a linear system of

equations Lesson Fifteen

Solving system of linear equations using elimination Checking whether a coordinate pair satisfies and equation Finding the coordinates of the point of intersection of two equations using algebra Recognizing the meaning of the point of intersection of a linear system of

equations



Unit Four Lesson Sixteen

Collecting like terms Distributive law Expanding second degree polynomial expressions Simplifying second degree polynomial expressions

Lesson Seventeen

Finding the greatest common factor Dividing polynomials

Lesson Eighteen

Factoring quadratic relations of the form where a = 1 cbxax 2 ++ Factoring difference of squares trinomials Factoring perfect square trinomials

Lesson Nineteen

Recognizing the direction of the opening of a parabola Recognizing the values of the maximum and minimum Recognizing the values of the x and y intercepts Recognizing the coordinates of the vertex Stating the equation of the axis of symmetry

Lesson Twenty

Understanding the meaning of “a” in the equation k)hx(ay 2 +−= Understanding the meaning of “h” and “k” in the equation k)hx(ay 2 +−= Finding the axis of symmetry given the an equation in vertex form

k)hx(ay 2 +−= Plotting and graphing quadratic equations Substitution into quadratic equations


Similar Triangles

Lesson 1

MFM2P – Foundations of Mathematics Unit 1 – Lesson 1

Lesson One Concepts

Similar triangle notation Interior angle property of similar triangles Proportionate side property of similar triangle



Similar Triangles

Similar triangles are triangles that have the same shape but not necessarily the same size.

For a two triangles to be considered similar either one of two properties must apply.

This is the standard notation used for showing two triangles are similar to one another ΔABC ≈ ΔXYZ. The order of the first three letters in not important, but what is important is that each of the first three letters corresponds with its matching letter from the other triangle.



Example 2 Prove that the following two triangles are similar and write the similarity statement .

Solution In ΔEFG ∠G = 180 - 65 - 35 = 80° and in ΔSTU ∠U = 180 - 80 - 35 = 65° Therefore both triangles have the same corresponding angles so ΔEFG ≈ ΔTUS

Support Questions 1. For each of the following pairs of triangles explain whether or not they are similar. Write a similarity statement for each pair of similar triangles.



Support Questions

2. For each of the following pairs of triangles explain whether or not they are similar. Write a similarity statement for each pair of similar triangles.

a. b.

Property 2. The ratios of corresponding sides are equal. Example 1

ZXCA

YZBC

XYAB

==

CA is the length from C to A

21224

ZXCA

27

14YZBC

236

XYAB

==

==

==

or

5.02412

CAZX

5.0147

BCYZ

5.063

ABXY

==

==

==

Since the ratios of corresponding sides are the same then the triangles are similar. ΔABC ≈ ΔXYZ



Example 2 Prove that the following two triangles are similar and write the similarity statement.

Solution

31133

UTGE

34

12SUFG

35.85.25

TSEF

==

==

==

or

33.031

3311

GEUT

33.031

124

FGSU

33.031

5.255.8

EFTS

===

===

===

Therefore both triangles have the same corresponding angles so ΔEFG ≈ ΔTSU

Support Questions 3. For each of the following pairs of triangles explain whether or not they are

similar. Write a similarity statement for each pair of similar triangles.



Support Questions 3. For each of the following pairs of triangles explain whether or not they are


Key Questions #1 1. For each of the following pairs of triangles explain whether or not they are




Key Questions #1 2. For each of the following pairs of triangles explain whether or not they are


3. For each of the following pairs of triangles explain whether or not they are


4. These triangles are the same shape yet have different sizes and orientations. Are the triangles similar? Explain.

5. The following pairs of triangles are similar. Determine the value of x and y in each.



Key Questions #1 6. In the diagram given below, a man 1.5 m tall is standing outside a church.

How tall is the church?


The Pythagorean Theorem

Lesson 2


Lesson Two Concepts

Triangle notation Substitution into the Pythagorean theorem Using Pythagorean theorem to find the missing side of a triangle



Pythagorean Theorem The Pythagorean Theorem is one method used to calculate an unknown side of a right angle triangle if the other two side are known.

Since the formula for the area of a square is then to find the length of the a side of a square we square root the value of the area.

2sA =

Example Find the length of a side of a square that has an area of 100 . 2cm Solution The area of square was 100 then the side would be 10.

10100 ±=

We only use the positive answer since you cannot have negative length.

Example a. Find the length of the hypotenuse in the given triangle using Pythagorean Theorem. 3m h 4m



b. Find the length of the missing side in the given triangle using Pythagorean Theorem. x 18 cm 12 cm Solution a. Using Pythagorean Theorem 222 cba =+

2

222

222

c25c43cba

=

=+

=+

5 = c

The area of square created by the hypotenuse is 25 so the length of its side is the square root of 25.

Therefore, the length of the hypotenuse is 5 m. b. Using Pythagorean Theorem 222 cba =+

180x1218x1812x

cba

2

222

222

222

=

−=

=+

=+

x ≈ 13.4 cm Therefore, the length of the missing side is 13.4 cm.

Support Questions 1. Calculate the length of the third side of each triangle. Round to one decimal place. a.

7 m 12 m



Support Questions 1. Calculate the length of the third side of each triangle. Round to one decimal place. b. 10.2 cm 14.7 cm 2. Calculate the diagonal of each rectangle. Round to one decimal place.

a. 8 cm 14 cm b. 1.5 m

2.7 m 3. The lengths of the sides of three triangles are given. Which are right triangles? a. AB = 5 m, BC = 6.5 m, CA = 8.02 m b. DE = 3 m, EF = 4 m, FD = 5 m c. GH = 7.6 m, HI = 3.6 m, IG = 11.0 m



Key Questions #2 1. What are the square roots of each number?

a. 14 b. 81 c. 100 d. 1

e. .625 f. 1.73 g. 251 h.

6436

2. Calculate the length of the third side of each triangle. Round to one decimal place. a.

5 m 12 m b. 10.1 cm 18.25 cm

3. The lengths of the sides of three triangles are given. Which are right triangles? a. AB = 24 m, BC = 10 m, CA = 26 m b. DE = 7 m, EF = 8 m, FD = 13 m c. GH = 10.6 m, HI = 5.6 m, IG = 9.0 m

4. A 7.5 m ladder is placed with its lower end 2 m away from the wall. How high up the wall does the ladder reach? 5. Noah rows across a river that is 48 m wide. The current carries her 15 m

downstream. How far does Noah actually travel?



Key Questions #2 6. A new television measures 24” by 18”. What is the length of the diagonal of

the television?

7. The length of each rafter for a roof is 60 m including a 1 m overhang. The peak is 20.5 m above the horizontal beam. Find the width of the horizontal beam.

8. An electrical pole is 12.4 m tall. It is supported by a guy wire that is 17.3 m long. How far is the anchor of the guy wire to the base of the electrical pole?


Sine Ratio

Lesson 3


Lesson Three Concepts

Label parts of a right triangle i.e. opposite, adjacent and hypotenuse Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse Using the sine ratio to find unknown sides of a right triangle Using the sine ratio to find unknown interior angles of a right triangle



Trigonometry: The Sine Ratio

The sine ratio can be used calculate an unknown angle or and unknown side in a right triangle. To find the either of these unknowns the opposite side and hypotenuse are used from an angle of reference. Example Suppose we use ∠A as the angle of reference then:

Sine always used the hypotenuse and opposite sides of a right triangle. Sine Opposite Hypotenuse: SOH When ∠ A is an acute angle in a right triangle, then

Sin A = side Hypotenuse of Length

Aangle Opposite side of Length

Sin A = ypotenuse

ppositeH

O



How to find the missing side of a right triangle using the sine ratio. Example 1 Find the value of x.

Solution

Sin A =HO



Support Questions 1. In each triangle, name the side: a. opposite ∠E b. the hypotenuse

2. Calculate the Sin A and Sin B in each triangle. a. b.

3. Calculate. a. Sin 35° b. Sin 71° c. Sin 55° d. Sin 90° 4. Calculate the value of x. a. b.



Support Questions 4. Calculate the value of x. c. d.

How to find the missing angle of a right triangle using the sine ratio. Example 2 Find ∠A.

Solution

Sin A =HO



Support Questions 5. Calculate.

a. Sin 0.725 b. Sin 0.325 c. Sin1− 1− 1−

73 d. Sin 1−

125

6. Calculate ∠E to the nearest degree.

a. Sin E = 0.625 b. Sin E = 0. 812 c. Sin E = 53 d. Sin E =

117

7. Calculate ∠E.



Key Questions #3 1. In each triangle, name the side:

2. Calculate the Sin A and Sin B in each triangle.

3. Calculate. a. Sin 42° b. Sin 68° c. Sin 12° d. Sin 0° 4. Calculate the value of x. a. b.



Key Questions #3 5. Calculate.

a. Sin 0.612 b. Sin 0.825 c. Sin1− 1− 1−

52 d. Sin 1−

133


a. Sin E = 0.387 b. Sin E = 0. 900 c. Sin E = 2912 d. Sin E =

135

7. Calculate ∠ D.

8. A guy wire is 13.5 m long. It supports a vertical power pole. The wire is fastened to the ground 9.5 m from the base of the a 8.7 m tall pole. Calculate the measure of the guy wire and the ground.



Key Questions #3 9. A 5.0 m ladder is leaning 3.7 m up a wall. What is the angle the ladder

makes with the ground? 10. A 12 m ladder leaning up against a wall makes a 50° angle with the ground.

How far up the wall does the ladder reach?

11. Explain why Sin A = 35 is an error. Draw and label a triangle with the

known side lengths given in the ratio and then explain why the error occurs.


Cosine Ratio

Lesson 4


Lesson Four Concepts

Label parts of a right triangle i.e. opposite, adjacent and hypotenuse Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse Using the cosine ratio to find unknown sides of a right triangle Using the cosine ratio to find unknown interior angles of a right triangle



Trigonometry: The Cosine Ratio

The cosine ratio can be used calculate an unknown angle or and unknown side in a right triangle. To find the either of these unknowns the adjacent side and hypotenuse are used from an angle of reference. Example Suppose we use ∠A as the angle of reference then:

Cosine always used the hypotenuse and adjacent sides of a right triangle. Cosine Adjacent Hypotenuse: CAH When ∠ A is an acute angle in a right triangle, then

Cos A = side Hypotenuse of Length

Aangle Adjacentside of Length

Cos A = ypotenuse

djacentH

A



How to find the missing side of a right triangle using the Cosine ratio. Example 1 Find the value of x.

Solution

Cos A =HA



Support Questions 1. In each triangle, name the side: a. adjacent ∠ E b. adjacent ∠ Y

2. Calculate the Cos A and Cos B in each triangle. a. b.

3. Calculate. a. Cos 35° b. Cos 71° c. Cos 55° d. Cos 90° 4. Calculate the value of x. a. b.




How to find the missing angle of a right triangle using the Cosine ratio. Example 2 Find ∠C.

Solution




a. Cos 0.725 b. Cos 0.325 c. Cos1− 1− 1−

73 d. Cos 1−

125


a. Cos E = 0.625 b. Cos E = 0. 812 c. Cos E = 53 d. Cos E =

117

8. Calculate ∠E.




2. Calculate the Cos A and Cos B in each triangle.

3. Calculate. a. Cos 42° b. Cos 68° c. Cos 12° d. Cos 0° 4. Calculate the value of x. a. b.




a. Cos 0.612 b. Cos 0.825 c. Cos1− 1− 1−

52 d. Cos 1−

133


a. Cos E = 0.387 b. Cos E = 0. 900 c. Cos E = 2912 d. Cos E =

135

7. Calculate ∠E.

8. For safety, the angle between a ladder and the ground should be between 60 and 78. A ladder 8 m in length is placed 3 m from the base of a wall. Is it safe to climb the ladder?



Key Questions #4 9. A kite has a string 100 m long anchored to the ground. The string makes and

angle with the ground of 68°. What is the horizontal distance of the kite from the anchor?

10. A ladder is leaned against a wall with its base 6 m from the wall. The ladder makes a 50° angle with the ground. How long is the ladder?

11. Does the cosine ratio work with non-right triangles? Explain and prove your answer with an example.


Tangent Ratio

Lesson 5


Lesson Five Concepts

Label parts of a right triangle i.e. opposite, adjacent and hypotenuse Recognize parts of a right triangle i.e. opposite, adjacent and hypotenuse Using the tangent ratio to find unknown sides of a right triangle Using the tangent ratio to find unknown interior angles of a right triangle



Trigonometry: The Tangent Ratio

The Tangent ratio can also be used calculate an unknown angle or and unknown side in a right triangle. To find the either of these unknowns the adjacent side and opposite sides are used from an angle of reference. Example Suppose we use ∠A as the angle of reference then:

Tangent always used the opposite and adjacent sides of a right triangle. Tangent Opposite Adjacent: TOA When ∠ A is an acute angle in a right triangle, then

Tan A = side Adjacentof Length

Aangle opposite side of Length

Tan A = djacentpposite

AO



How to find the missing side of a right triangle using the Tangent ratio. Example 1 Find the value of x.

Solution

Tan A =HA



Support Questions 1. In each triangle, name the side: a. adjacent ∠ F b. opposite ∠ Y

2. Calculate the Tan A and Tan B in each triangle. a. b.

3. Calculate. a. Tan 35° b. Tan 71° c. Tan 55° d. Tan 90° 4. Calculate the value of x. a. b.




How to find the missing angle of a right triangle using the Cosine ratio. Example 2 Find ∠A.

Solution




a. Tan 0.725 b. Tan 0.325 c. Tan1− 1− 1−

73 d. Tan 1−

125


a. Tan E = 0.625 b. Tan E = 0. 812 c. Tan E = 53 d. Tan E =

117

8. Calculate ∠E.




2. Calculate the Tan A and Tan B in each triangle.

3. Calculate. a. Tan 42° b. Tan 68° c. Tan 12° d. Tan 50° 4. Calculate the value of x. a. b.




a. Tan 1.612 b. Tan 0.825 c. Tan1− 1− 1−

25 d. Tan 1−

133


a. Tan E = 0.387 b. Tan E = 2 .90 c. Tan E = 2912 d. Tan E =

513

7. Calculate ∠E.

8. Building A and building B are 15 m apart. building B is 75 m high. What is the angle (angle of elevation) from the base of building A to the top of building B?



Key Questions #5 9. The top of a communications tower is 150 m above sea level. From a boat at sea, its angle of elevation is 3°.

a. Using the diagram given above, what is meant by the angle of elevation? b. How far is the boat from the tower? 10. A ladder is leaned 10 m up a wall with its base 6 m from the wall. What angle does the ladder make with the ground?

11. The acronym SOHCAHTOA is often used in trigonometry. What do you think each letter stands for and give an example finding either an unknown side or

an unknown angle using a portion of this acronym.


MFM2P – Foundations of Mathematics Unit 1 - Support Question Answers

Answers to Support Questions: Unit One Lesson One 1a. similar; all angles correspond; ΔDEF≈ΔABC b. similar; all angles correspond; ΔQRS≈ΔTUV c. not similar because don’t know if all angles correspond with each other d. similar; all angles correspond; ΔABC≈ΔDEF 2a. similar; all angles correspond; ΔABC≈ΔDEF b. similar; all angles correspond; ΔQRS≈ΔXWY

3a. similar; sides are proportionate; 236

24

12

===

b. similar; sides are proportionate; 5525

420

315

===

c. not similar; sides are not proportionate; 96

63

42

≠=

d. not similar; sides are not proportionate; 1235

515

39

≠=

Lesson Two 1a. b.

c9.13c193

c14449c127

2

2

222

≈=

=+

=+

6.10a05.112a

04.10409.216a)2.10()7.14(a

)2.10()7.14()2.10()2.10(a)7.14()2.10(a

2

2

222

22222

222

≈=

−=

−=

−=−+

=+

2a. b.

c1.16c260

c19664c148

2

2

222

≈=

=+

=+

c1.3c54.9

c29.725.2c)7.2()5.1(

2

2

222

≈=

=+

=+



3a. b. c.

no3204.6425.67

3204.6425.422502.85.65 222

≠=+

=+

yes2525

25169543 222

==+=+

no12172.70

12196.1276.57116.36.7 222

≠=+

=+

Lesson Three 1a. GE b. YZ

2a. 47.3SinA = ;

45.1SinB = b.

1812SinA = ;

184.13SinB =

3a. 0.574 b. 0.946 c. 0.819 d. 1 4a. b. c. d.

cm23.2x

37Sin7.3x7.3

x37Sin

==

=

cm03.7x

23Sin18x18x23Sin

==

=

x28.5

x48Sin

448Sin48Sinx

48Sin4

48Sinx4x448Sin

=

=

=

=

=

x2.15

x62Sin4.13

62Sin62Sinx

62Sin4.13

62Sinx4.13x

3.1362Sin

=

=

=

=

=

5a. 46° b. 19° c. 25° d. 25°

6a. b. c. °=

−

39625.0Sin 1

°=

−

54812.0Sin 1

°=

−

3753Sin 1

d. °=

−

40117Sin 1

7a. b. c. d.

°=∠

=∠

=

−

35E7.31.2SinE

7.31.2SinE

1

°=∠

=∠

=

−

34E1810SinE

1810SinE

1

°=∠

=∠

=

−

26E94SinE

94SinE

1

°=∠

=∠

=

−

56E1.164.13SinE

1.164.13SinE

1



Lesson Four 1a. GE b. XY

2a. 45.1CosA = ;

47.3CosB = b.

184.13CosA = ;

1812CosB =

3a. 0.819 b. 0.326 c. 0.574 d. 0 4a. b. c. d.

cm95.2x

37Cos7.3x7.3

x37Cos

==

=

cm57.16x

23Cos18x18x23Cos

==

=

x98.5

x48Cos

448Cos48Cosx

48Cos4

48Cosx4x448Cos

=

=

=

=

=

x5.28

x62Cos4.13

62Cos62xCos

62Cos4.13

62Cosx4.13x

3.1362Cos

=

=

=

=

=

5a. 44° b. 71° c. 65° d. 65°

6a. b. c. °=

−

51625.0Cos 1

°=

−

36812.0Cos 1

°=

−

5353Cos 1

d. °=

−

50117Cos 1

7a. b. c. d.

°=∠

=∠

=

−

55E7.31.2CosE

7.31.2CosE

1

°=∠

=∠

=

−

56E1810CosE

1810CosE

1

°=∠

=∠

=

−

64E94CosE

94CosE

1

°=∠

=∠

=

−

34E1.164.13CosE

1.164.13CosE

1



Lesson Five 1a. FG b. XZ

2a. 5.17.3TanA = ;

7.35.1TanB = b.

4.1312TanA = ;

124.13TanB =

3a. 0.700 b. 2.90 c. 1.43 d. undefined (error) 4a. b. c. d.

cm79.2x

37Tan7.3x7.3

x37Tan

==

=

x4.42

x23Tan

1823Tan23Tanx

23Tan18

23Tanx18x

1823Tan

=

=

=

=

=

m44.4x

48Tan4x4x48Tan

==

=

m2.25x

62Tan4.13x4.13

x62Tan

c

==

=

5a. 36° b. 18° c. 23° d. 23°

6a. b. c. °=

−

32625.0Tan 1

°=

−

39812.0Tan 1

°=

−

3153Tan 1

d. °=

−

32117Tan 1

7a. b. c. d.

°=∠

=∠

=

−

60E1.27.3TanE

1.27.3TanE

1

°=∠

=∠

=

−

61E1018TanE

1018TanE

1

°=∠

=∠

=

−

66E49TanE

49TanE

1

°=∠

=∠

=

−

50E4.131.16TanE

4.131.16TanE

1


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