MCR 3U Functions Grade 11
Trigonometry 11 Test 3
K/U
APP
TIPS
COM
𝟏𝟐
𝟖
𝟏𝟓
𝟏𝟕
1. Determine the values of 𝜃 if cot 𝜃 = 2 and 0o ≤ 𝜃 ≤ 360o. [K/U ]
2
Solution
cot 𝜃 = 2 ⟹ tan 𝜃 = 0.5
tan 𝜃 > 0 ∴ 𝜃 is in the I Quadrant or 𝜃 is in the III Quadrant.
Related angle 𝛼 = tan– 1 (0.5) ≈ 26.6o.
Quadrant I
Quadrant III
𝜃 = 𝛼 = 26.6o,
𝜃 = 180o + 𝛼 = 180o + 26.6o = 206.6o.
2. Without using a calculator, determine the exact value of
cos 60o ∙ sin 210o.
[APP 2 ]
Solution
1
cos 60o ∙ sin 210o = sin (180o + 30o)
2
1
1
2
2
= (– sin 30o) =
1
1
2
4
(– ) =– .
3. A boat is approaching a cliff which is 50 m tall. If the angle of elevation
from the boat is 600 , how far away is the boat from the cliff? Give the
exact value, not an approximation. [APP ]
3
Solution
50
𝑥
= tan 60o ⟹
50
𝑥
= √3 ⟹ x =
50
√3
=
50√3
3
m.
4. An angle 𝜃 lies in quadrant 3.
a). Without using a calculator, determine which ratio is valid. [T ]
2
i. sin 𝜃 = 0.798
iii. sin 𝜃 = – 1.083
ii. sin 𝜃 = – 0.573
iv. tan 𝜃 = 0.841
b). Explain how you arrived at your answer in a). [C ]
2
Solution
𝜃 lies in quadrant 3 ∴ sin 𝜃 < 0 and tan 𝜃 < 0 ∴ not “i.” and not “iv.”
– 1 ≤ sin 𝜃 ≤ 1 ∴ not “iii.” Therefore, the right choice is “ii.”: sin 𝜃 = –
0.573
5. The point P (2, –3) lies on the terminal arm of an angle in standard
position. What is the value of the principle angle 𝜃?
Solution
𝑦
3
x = 2, y = –3 ⟹ tan 𝜃 = = – .
Related angle 𝛼 = tan
𝑥
3
–1 ( )
2
[APP 2 ]
2
≈ 56o.
𝜃 = 360o – 56o = 304o
6. An angle 𝜃 is the principle angle that lies in quadrant 4 such that cos 𝜃
2
= , 0° ≤ 𝜃 ≤ 360°. Determine the exact values of x, y, r and 𝜃 to the
5
nearest degree. [K/U ]
2
Solution
2
cos 𝜃 = ∴ x = 2, r = 5;
5
2
2
𝑥 + 𝑦 = 𝑟 2 ⟹ 22 + 𝑦 2 = 52 ⟹ 4 + 𝑦 2 = 25 ⟹ 𝑦 2 = 21.
𝜃 lies in quadrant 4 ∴ y < 0 ⟹ y = –√21.
2
Determine the related angle: 𝛼 = cos– 1 ( ) ≈ 66°.
5
Since 𝜃 is the principle angle that lies in quadrant 4, we get:
𝜃 = 360° – 66° = 294°.
7. Given:
tan 𝜃
√3
= 1, 0° ≤ 𝜃 ≤ 360°.
a) Without using a calculator, determine 𝜃 .
[T 2 ]
Solution tan 𝜃 = √3
Related angle 𝛼 = tan– 1 (√3) = 60°.
tan 𝜃 > 0 ∴ 𝜃 lies in quadrant I or 𝜃 lies in quadrant III.
Quadrant I
Quadrant III
𝜃 = 𝛼 = 60°.
𝜃 =180° + 𝛼 = 240°.
b) Explain what you did to get your answer for a). [C
3
]
Solution
There exist two principal angles 𝜃 such that tan 𝜃 = √3. One angle (𝜃 =
60°) lies in the I Quadrant, and the other angle (𝜃 = 240°) lies in the III
Quadrant.
8. Simplify: (1 – cos 𝜃) (1 + cos 𝜃). [K/U ]
2
Solution
(1 – cos 𝜃) (1 + cos 𝜃) = 1 – cos2 𝜃 = sin2 𝜃.
9. What step is missing in the following proof of the identity
cos2 𝜃
1 − sin 𝜃
cos2 𝜃
1 – sin 𝜃
= 1 + sin 𝜃 ? [T ]
3
= missing step =
(1 − sin 𝜃)(1 + sin 𝜃)
1 – sin 𝜃
= 1 + sin 𝜃 .
Answer: Missing step = cos2𝜃 = 1 – sin2𝜃
10. Determine the measure of the angle 𝜃 in the diagram below: [K/U
Solution
By Sine Law:
𝐵𝐶
sin 𝐴
17.9 sin 41°
=
𝐴𝐵
sin 𝐶
⟹
17.9
sin 𝜃
=
12.2
sin 41°
sin 𝜃 =
≈ 0.9626
12.2
sin– 1 (0.9626) ≈ 74° ⟹ 𝜃 ≈ 180° – 74° = 106°.
2
]
11. Determine ∠𝐵 for the triangle with a = 3.7, b = 4.9, and ∠ A = 46°.
[K/U 3 ]
Solution
ℎ
From ∆ ACH: sin 46o = ⟹
4.9
o
h = 4.9 sin 46 ≈ 3.5 m
Since b = AC = 4.9 m,
we have: h < a < b ∴ 2 triangles.
From ∆ ABC by Sine Law:
𝐵𝐶
𝐴𝐶
3.7
4.9
=
⟹
=
⟹
sin∠ 𝐴
sin ∠𝐵
sin 46°
sin ∠𝐵
4.9 sin 46o
sin ∠B =
≈ 0.9526.
3.7
∠B = sin– 1 (0.9526) ≈ 72o or A = 180o – 72o = 108o.
Case 1: ∠AB1C = 72o, ∠ACB1 = 180o – 46o – 72o = 62o.
From ∆AB1C by Sine Law:
𝐴𝐵1
sin ∠𝐴𝐶𝐵1
=
𝐵1 𝐶
sin ∠𝐴
⟹
𝐴𝐵1
sin 620
=
3.7
sin 46o
⟹ AB1=
3.7 sin 62o
sin 460
= 4.54 m
Case 2: ∠ AB2C = 108o, ∠ ACB2 = 180o – 46o – 108o = 26o
From ∆ AB2C by Sine Law:
𝐴𝐵2
sin 𝐴𝐶𝐵2
=
𝐵2 𝐶
sin ∠𝐴
⟹
𝐴𝐵2
sin 260
Answer: 2.25 m, 4.54 m.
=
3.7
sin 46o
⟹ AB2=
3.7 sin 26o
sin 460
= 2.25 m
12. A triangular lot has one side 450 m long, the opposite angle 680 , and
the adjacent side 200 m long. Determine the length of the third side.
[APP 2 ]
Solution
Since 450 > 200, there is one solution, and
∠C is an acute angle. By Sine Law,
450
200
200 sin 68°
=
⟹ sin ∠C =
sin 68°
sin ∠𝐶
450
450 > 250 ⟹ ∠C < ∠A= 68°⟹∠C is acute.
sin ∠C≈ 0.4121⟹ ∠C≈ sin– 1 (0.4121) ≈24°
∠B = 180° – 68° – 24° ≈ 88°.
By Sine Law,
𝐴𝐶
450
450 sin 88°
=
⟹AC =
≈ 485 m.
sin 88°
sin 68°
sin 68°
13. Find x in the diagram. [K/U ]
2
Solution By Cosine Law,
x2 = 4.62 + 6.32 – 2(4.6) (6.3) cos 77°
x2 ≈47.8
x ≈ 6.9 m.
14. In ∆ ABC, a = 6, b = 16, ∠C = 60°. Find ∠𝐵. [K/U
2
]
Solution By Cosine Law, c2 = a2 + b2 – 2ab cos ∠C. Hence,
c2 = 62 + 162 – 2(6) (16) cos ∠60° = 196 ⟹ c= 14.
By Sine Law,
𝑏
𝑐
16
14
16 sin 60°
=
⟹
=
⟹ sin ∠ B =
≈ 0.9897.
sin ∠𝐵 sin ∠𝐶
sin ∠𝐵 sin 60°
14
∠ B = sin – 1 (0.9897) ≈ 82° or ∠ B = 180° – 82° = 98°. Ambiguous case!
Therefore, it is better to use the Cosine Law:
cos ∠ B =
𝑎2 + 𝑐 2 − 𝑏2
2𝑎𝑐
=
62 + 142 − 162
2(6)(14)
∠ B = cos – 1 (– 0.143) ≈ 98°.
= – 0.143
15. Prove the identity: tan2x ∙ cos2x =
Solution
(sec2 𝑥 − 1) ∙ (1 − sin4 𝑥)
1 + sin2 𝑥
[T 5 ]
(1 + tan2 𝑥 − 1) ∙ (1 − sin2 𝑥)(1 + sin2 𝑥)
1 + tan2 𝜃 = sec 2 𝜃
R.S. =
2
1 + sin 𝑥
2
= tan x ∙ (1 – sin2 x) = tan2x ∙ cos2x = R.S.
16. Ben and Cara are both looking at the top of a tower in the distance in
the same line of sight, although Cara is 100 m ahead of Ben. Ben observed
the tower at an angle of 16° with the ground. If the top of the tower is 800
m from Cara, what is Cara’s angle of sight with the ground? First make a
sketch and then solve. [A C ]
5
1
Solution
From ∆ ABC by Sine Law:
𝐵𝐶
𝐴𝐶
=
⟹
sin ∠𝐵𝐴𝐶
100
sin ∠𝐵𝐴𝐶
=
sin ∠𝐴𝐵𝐶
800
⟹
sin 16°
100 sin 16°
sin ∠BAC =
≈ 0.034454 ⟹
800
∠BAC = sin– 1(0.034454) ≈ 2°.
By Exterior Angle Theorem, we get:
𝜃 = ∠ACD = ∠ABC + ∠BAC = 16° + 2° = 18°.
17. Explain why tan 90° and tan 270° are undefined. [C ]
2
Solution
The 90°-angle is represented by the point
(0, 1) on the unit circle. Hence, x = 0, y = 1,
𝑦
1
and tan 90° = = – undefined.
𝑥
0
The 270°-angle is represented by the point
(0, –1) on the unit circle. Hence, x = 0,
𝑦
–1
𝑥
0
y = –1, and tan 270° = =
– undefined.