Unit 1 Trig.notebook · Unit 1 Trig.notebook 5 February 04, 2013 Sep 64:35 PM Example 2: Solve for...
Transcript of Unit 1 Trig.notebook · Unit 1 Trig.notebook 5 February 04, 2013 Sep 64:35 PM Example 2: Solve for...
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Unit 1 Trig.notebook
1
February 04, 2013
Sep 79:27 AM
Area 1
Area 2
Area 3
Can you find a relationshipwith the three areas?
Bell Work
Sep 64:21 PM
Unit 1:Trigonemetry
Key Ideas:
• Pythagorean Theorem• Primary trig ratios: sine, cosine & tangent• Inverses: sin1, cos1, tan1• Obtuse angles • Solving Triangles: SOHCAHTOA, Sine Law, Cosine Law• Applications: Word Problems
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Unit 1 Trig.notebook
2
February 04, 2013
Sep 64:19 PM
Day 1: Pythagorean Theorem:
c2 = a2 + b2
• always "c"• opposite the right angle• the longest side
WARNING: When solving for one of the other two sides use:
a2 = c2 b2 b2 = c2 a2
They mean the same thing!
hypotenuse
Sep 64:23 PM
Examples: Solve for the missing side. Round to one decimal place.
108
x
2.3
3.7
x
x2 = 102 82
a2 = c2 b2
x2 = 100 64
x2 = 36
x = √36x = 6
c2 = a2 + b2
x2 = 2.32 + 3.72
x2 = 5.29 + 13.69
x2 = 18.98
x = √19.98x ≈ 4.4
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Unit 1 Trig.notebook
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February 04, 2013
Sep 64:25 PM
Day 2: Primary Trigonometric Raos: Sine, Cosine & Tangent
Example 1: Evaluate to 3 decimal places.
sin 650 = cos 1240 = tan 3410 =
angles
Example 2: Solve for the angle to the nearest degree.
sin R = 0.25 cos B = 0.92 tan Q = 1.54
Recall: When solving for an angle we must use the inverse functions!
R = sin1(0.25) B = cos1(0.92) R = tan1(1.54)
0.906 0.559 0.344
R = 14.50 B = 23.10 R = 57.00
Sep 64:32 PM
When do we use these?
When we want to solve for an angle or a side in a right angle triangle.
A
BC
What is the first thing that we must do when solving a right angle triangle?
LABEL THE SIDES: OPPOSITE, ADJACENT, HYPOTENUSE
hypostenuse
adjacent
opposite
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Unit 1 Trig.notebook
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February 04, 2013
Sep 64:34 PM
A
BC
Recall: SOHCAHTOA
Hence, the primary trig ratios for angle A below are:
hypostenuseadjacent
opposite
sin A =
opposite
hypotenusecos A =
adjacent
hypotenusetan A =
adjacent
opposite
Sep 64:34 PM
Example 1: Solve for side b.
420
12 cm
b
opposite
adjacenthypotenuse
sin 420 = opposite
hypotenuse
sin 420 = b
12
b = 12sin420 (12 x sin420)
b = 8.0 cm
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Unit 1 Trig.notebook
5
February 04, 2013
Sep 64:35 PM
Example 2: Solve for side p.
670
9.7 cm
p
oppositeadjacent
hypotenuse
tan 670 = opposite
adjacent
tan 670 = 9.7
p
p = 9.7tan 670
(9.7 ÷ tan670)
p = 4.1 cm
Sep 98:11 AM
Bell WorkSolve for side b in the following two triangles.
230
11 in
b
5 cm
3 cm
b
sin 230 =
opp
adj
hyp
opp
hyp
b
11
b = 11sin230
sin 230 =
b ≈ 4.3 in
b2 = c2 a2
b2 = 52 32
b2 = 25 9
b2 = 16
b = √16
b = 4 cm
1
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Unit 1 Trig.notebook
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February 04, 2013
Sep 98:27 AM
Day 3: Solving for a missing angle in a right angle triangle
Example 1: Evaluate to 3 decimal places.
sin 650 = cos 1240 = tan 3410 =
angles
Example 2: Solve for the angle to the nearest degree.
sin R = 0.25 cos B = 0.92 tan Q = 1.54
Recall: When solving for an angle we must use the inverse functions!
R = sin1(0.25) B = cos1(0.92) R = tan1(1.54)
0.906 0.559 0.344
R = 14.50 B = 23.10 R = 57.00
Recall from yesterday our primary trig ratios: sine, cosine and tangent
Sep 98:30 AM
When do we use these?
When we want to solve for an angle or a side in a right angle triangle.
A
BC
What is the first thing that we must do when solving a right angle triangle?
LABEL THE SIDES: OPPOSITE, ADJACENT, HYPOTENUSE
hypostenuse
adjacent
opposite
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Unit 1 Trig.notebook
7
February 04, 2013
Sep 98:32 AM
A
BC
Recall: SOHCAHTOA
Hence, the primary trig ratios for angle A below are:
hypostenuseadjacent
opposite
sin A = opposite
hypotenuse
cos A = adjacent
hypotenuse
tan A = adjacent
opposite
Sep 98:32 AM
Example 1: Solve for angle Q.
11 m
opposite
adjacent
hypotenuse
cos Q = hypotenuse
adjacent
cos Q = 11
17
Q = (cos1 (11÷17) )
Q = 49.70
Q
RS
17 m
11
17cos1 [ [
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Unit 1 Trig.notebook
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February 04, 2013
Sep 98:33 AM
Example 2: Solve for angle D.
8.7 inopposite
adjacenthypotenuse
tan D = opposite
adjacent
tan D = 8.7
12.1
D = (tan1 (8.7÷12.1) )
D ≈ 35.70
E F
D
12.1 in
tan1 [ [8.712.1
Sep 910:23 AM
Bell WorkSolve for the indicated variable.
15 cm
8 cmx
5.1 in
9.7 in
X
Tool:Pythagorean Theorem
Solution:
c2 = a2 + b2
x2 = 82 + 152
x2 = 64 + 225
x2 = 289x = √289x = 17 cm
Pythagorean TheoremLongest Side
Tool: SOHCAHTOA
Inverse (sin1, cos1, tan1)
Solution:
tan X = opp
adj
tan X = 9.75.1
opp
adj
hyp
X = tan1( )9.75.1X ≈ 62.30
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Unit 1 Trig.notebook
9
February 04, 2013
Sep 112:57 PM
Day 4: Sine, Cosine, and Tangent of Obtuse Angles
Recall:
• An ACUTE ANGLE is less than 900.
• A RIGHT ANGLE is equal to 900.
• An OBTUSE ANGLE is bigger than 900 and less than 1800.
Sep 113:05 PM
THE CARTESIAN GRID: Quadrant I ACUTE ANGLES (Less than 900)
Quadrant IQuadrant II
X
Quadrant III Quadrant IV
Summary: In Quadrant I (ACUTE ANGLES) sine, cosine and tangent are ALL POSITIVE!
opp
adj
hyp
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Unit 1 Trig.notebook
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February 04, 2013
Sep 113:18 PM
THE CARTESIAN GRID: Quadrant II OBTUSE ANGLES (900 1800)
Quadrant IQuadrant II
X
Quadrant III Quadrant IV
Summary: In Quadrant II (OBTUSE ANGLES) sine is POSITIVE
cosine and tangent are NEGATIVE!
opp
adj
hyp
Sep 113:03 PM
R = 570In Summary,
Sin A Cos A Tan AFor Acute angles (0°
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Unit 1 Trig.notebook
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February 04, 2013
Sep 113:24 PM
Example 1: State whether the following angle is obtuse, acute or either.
a) sin A = 0.2545 b) cos A = 0.9684 c) tan A = 1.5684
Either Acute Obtuse
Example 2: Solve for the following OBTUSE angle.
a) sin A = 0.2545 b) cos B = 0.5468 c) tan C = 0.2547
A = 1800 sin1(0.2545)
A ≈ 1800 14.70
A ≈ 165.30
B = cos1(0.5468)
B ≈ 123.10 C = 1800 [tan1(0.2547)]
C ≈ 1800 14.30
C ≈ 165.70
Sep 113:39 PM
Question 1: State whether the following angle is obtuse, acute or either.
a) sin A = 0.9657 b) cos A = 0.9562 c) tan A = 0.8564
Either AcuteObtuse
Question 2: Solve for the following OBTUSE angle.
a) sin A = 0.5812 b) cos B = 0.2568 c) tan C = 1.7854
A = 1800 sin1(0.5812)
A ≈ 1800 15.50
A ≈ 144.50
B = cos1(0.2568)
B ≈ 104.90 C = 1800 [tan1(1.7854)]
C ≈ 1800 60.70
C ≈ 119.30
Bell Work
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Unit 1 Trig.notebook
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February 04, 2013
Sep 113:47 PM
Can we use the Pythagorean Theorem? NO
Can we use SOHCAHTOA? NO
Then ... what can we use?
• Sine Law • Cosine Law
Day 5: Solving triangle that are NOT right angles ‐ THE SINE LAW
Sep 113:49 PM
Before we begin let's recall a few things about triangles:
#1 Sum of the angle in a triangle:
A
B
C
A + B + C = 1800
#2 Labelling Conventions:
P
Q
R
pr
• Angles are labelled with
• Sides are labelled with
upper case letters
lower case letters
• sides and angles correspond.Opposite
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Unit 1 Trig.notebook
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February 04, 2013
Sep 113:49 PM
Sine Law:
A
B
C
a
b
c a = b = c sin A sin B sin C
OR
sin A = sin B = sin C a b c
HINT: Look for the OPPOSITE SIDEANGLE PAIR!
If you are given two angles, always start by finding the third angle (i.e. 1800 minus the other two)
i.e. Are you given A & a, B & b or C & c?
Sep 113:51 PM
A
B
C
a
b
12
320
650
Example 1:
A = 1800 320 650 = 830
a = c sin A Sin C
Find the value of side a in the triangle below.
a = 12 sin 83 0 sin 65 0
12sin83 sin65 0
a = 0
a = 13.14
The size of opposite sideangle pairs should correspond (i.e. the biggest side should be across from the biggest angle).
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Unit 1 Trig.notebook
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February 04, 2013
Sep 113:54 PM
A
B
C
2.3
b
3.1
470
Example 2:
sin C = sin A c a
Find the value of angle C in the triangle below.
3.1sin47 2.3
sin C = 0
sin C = sin 47 3.1 2.3
0
3.1sin47 2.3
0C = sin-1 [ [
C = 80.30
Sep 113:55 PM
Bell Work
Question 1: How do we know to use the Sine Law?
Question 2: Solve for the missing side in the following triangles.
Not a right angle triangle given a side/angle pair.
350
620
x
15.1 cm
x570 15.1 cm
1800 350 620 = 830
830
x
sin83015.1
sin350=
x = 15.1sin830
sin350
x ≈ 26.1 cmo
ha
SOHCAHTOA
cos570 = x
15.1
x = 15.1cos570
x ≈ 8.2 cm
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Unit 1 Trig.notebook
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February 04, 2013
Sep 113:54 PM
Can we use the Pythagorean Theorem? NO
Can we use SOHCAHTOA? NO
Then ... what can we use?
• Sine Law • Cosine Law
Day 6: Solving triangle that are NOT right angles ‐ THE COSINE LAW
Sep 113:57 PM
Cosine Law:
A
B
C
a
b
c
a2 = b2 + c2 - 2bc cosA .OR
cosA = a2 + b2 - c2
2bc
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Unit 1 Trig.notebook
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February 04, 2013
Sep 113:58 PM
Example 1: Find the value of side b in the triangle below.
A
B
C
1411
230
b
b2 = 142 + 112 - 2(14)(11)cos230
b2 = 142 + 112 - 2(14)(11)cos230
b2 = a2 + c2 - 2ac cosB .
b = √(142 + 112 - 2(14)(11)cos230)
b ≈ 5.8
Sep 113:58 PM
Example 2: Find the value of angle A in the triangle below.
A
B
C
26
15
19
cosA = a2 + b2 - c2
2bc
cosA = 192 + 152 - 262
2(19)(15)
A = cos-1 192 + 152 - 262
2(19)(15)[ [
A = cos-1 -90
570[ [
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Unit 1 Trig.notebook
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February 04, 2013
Sep 115:18 PM
Bell Work
Question 1: How do we know to use the Cosine Law?
Question 2: Solve for the missing angle in the following triangles.
Not a right angle triangle can't use sine law last resort :)
X
9.7 cm
15.1 cm
12.3 cm 9.7 cm
X15.1 cm
9.72 + 15.12 12.32
2(9.7)(15.1)
o
ha
SOHCAHTOAcosX =
9.7
15.1
cosX ≈ 0.642
X ≈ cos-1(0.642)
Not a right angle triangle SAS sandwich OR SSS
X ≈ 500
cosX =
cosX = 170.81292.94
cosX ≈ 0.583
X ≈ cos-1(0.583)
X ≈ 540
Sep 153:11 PM
Example:
DRAW A PICTURE!
Two planes took off from Pearson International Airport at the same time. The first plane heads due west at an angle of elevation of 25o and a speed of 168 Km/hr. The second plane heads due east at an angle of elevation of 32o and a speed of 156 Km/hr. How far apart are the planes after 3 hours?
Plane 1 Plane 2
Airport
d
504 4681230
1800 250 320 = 1230
SIMPLIFY YOUR PICTURE!
Plane 1 Plane 2
Airport
250 320
d
DECIDE WHETHER TO USE SOHCAHTOA, SINE LAW OR COSINE LAW!
COSINE LAW
∴ the planes are 282.8 km apart after 3 hours.
168 km/hr x 3 hr = 504 km 156 km/hr x 3 hr = 468 km
d2 = 5042 + 4682 2(504)(468)cos1230
d2 = 5042 + 4682 2(504)(468)cos1230
d2 ≈ 729,970.2d ≈ √729,970.2d ≈ 282.8 km
Applications
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Unit 1 Trig.notebook
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February 04, 2013
Sep 153:32 PM
A
B
C230
21
14
Bell Work/Test PreparationTIPS: Sovle for angle B to the nearest degree.
x
x2 = 212 + 142 2(21)(14)cos230
x2 ≈ 95.7x ≈ √95.7x ≈ 9.89.8
sinB14
sin230
9.8=
sinB14sin230
9.8=
sinB ≈ 0.558
B ≈ sin1(0.558)
B ≈ 400
Sep 153:47 PM
Application Review Test Preparation
Jamie's house is 15 Km away from Becky's house and 9 Km away from Kevin's house. From his posion, Becky's and Kevin's house are separated by an angle of 70 0. How far apart, to the nearest kilometer are Becky's and Kevin's houses? (Draw a diagram)
Example 1:
Remember: Keep the picture simple!
J
B K
15 Km 9 Km700
x
x2 = 152 + 92 2(15)(9)cos700
x2 ≈ 213.65x ≈ √213.65
x ≈ 15
∴ Becky's and Kevin's houses are approx. 15 Km apart.
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Unit 1 Trig.notebook
19
February 04, 2013
Sep 208:21 AM
Application Review Test Preparation
Two cabins A and B are separated by a large pond. To determine the distance between the cabins, Beth walks 2 km from A to a point K. She measured angle A equal to 230 and angle K equal to 560. Determine the distance between the cabin, correct to 2 decimal places. (Draw a diagram)
Example 2:
Remember: Keep the picture simple!
pond
A B
K
x230
560
1010
1800 230 560 = 10102 Km
x
sin5602
sin1010=
x = 2sin560
sin1010
x ≈ 1.69 cm
∴ The two cabins are approx. 1.69 km apart.
Sep 208:51 AM
Trigonometry Decision Tree
Is it a rightangle triangle?
YES NO
Are you solving for a side?
YES Solving for an angle!
Use: SOHCAHTOA Inverses (sin1, cos1, tan1)
NO
Do you HAVE two sides?
Pythagorean Theorem
YES
c2 = a2 + b2 a2 = c2 b2
hypotenuse other leg
ADD SUBTRACT
NO
SOHCAHTOALabel Carefully!
Variable on the...
top
bottom
MULTIPLY DIVIDE
x = 14sin230 x = 14÷sin230
Do you have a SIDEANGLE PAIR?
Sine Law
Solving for a ...
side angle
xsinX
ysinY
=
xsinX
ysinY
=
INVERSE!
Cosine Law
x2 = a2 + b2 2ab cosX.
side
Solving for a ...
cosX = a2+b2x2
2ab
INVERSE!
angle
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