MCR3U Grade 11 University
Trigonometry Test 8
20K
10C
12A
12T
Com. Level
All answers should be accurate to 1 decimal place unless otherwise
stated. Don't forget to include units.
Knowledge amd Understanding [20 marks]
Multiple Choice Enter the letter of the correct answer for questions 1-9
into the table below.
1
2
3
4
5
6
7
8
9
9K
1. Which of the following trigonometric ratios is positive in the 2nd
quadrant?
a) sin 𝜃
b) cos 𝜃
c) tan 𝜃
d) sec 𝜃
Answer: a)
2. A co-terminal angle can be obtained by doing the following operation
to the original angle:
a) Adding 180o
b) Subtracting 720o
c) Multiplying by 360o
d) None of the options
Answer: b)
3. How many angles between 0o and 360o satisfy the equation?
1
cos 𝜃 = −
√2
a) 0 angles
b) 1 angle
c) 2 angles
d) 4 angles
Answer: c)
4. Which of the following expressions are equal to tan 𝜃?
a) tan(180o − 𝜃 )
b) tan(180o + 𝜃 )
c) tan(270o − 𝜃 )
d) tan(360o − 𝜃 )
Answer: b)
5. Given that ∠𝐵 = 90o , 𝑎 = 50 𝑚, and 𝑏 = 32 𝑚, how many triangles
are possible?
a) 0 triangles b) 1 triangle
c) 2 triangles d) 3 triangles
Answer: a)
6. If the initial arm is in standard position and the terminal arm is on the
negative y-axis, which angle is created?
a) 0o
b) 90o
c) 180o
d) 270o
Answer: d)
7. Which of the options below is equivalent to: csc 𝜃 ∙ tan 𝜃?
cos 𝜃
a) cos 𝜃
b) sec 𝜃
c) sin 𝜃
d) 2
sin 𝜃
Answer: b)
8. Which of the options below is equivalent to: sec 2 𝜃 ∙ (sin2 𝜃 − 1)?
a) tan2 𝜃
b) cot 2 𝜃
c) 1
d) – 1
Answer: d)
9. The point (x, y), which is on the unit circle, can be written as:
a) (cos 𝜃, sin 𝜃) b) (sin 𝜃, cos 𝜃) c) (cos x, sin y) d) (sin x, cos y)
Answer: a)
Full Solution: Show all your work for the following questions.
10. Determine the exact value of the following expression:
cos 45° ∙ csc 45° – sin 60° ∙ sec 60°
(Hint: use the special triangles). [3K]
Solution
1
1
cos 45° = , csc 45° =
= √2,
2
sin 45°
sin 60° =
√
√3
2
, sec 60° =
1
cos 60°
= 2.
cos 45° ∙ csc 45° – sin 60° ∙ sec 60° =
1
√2
∙ √2 –
√3
2
∙ 2 = 1 – √3.
Answer: 1 – √3.
4
11. Given that cos θ = – , and 0° < θ < 360°, find the values of θ to the
5
nearest degree. [4K]
Solution
Determine the related angle:
4
𝛼 = cos – 1 ( ) ≈ 37°
5
cos θ < 0 ∴ θ lies in the II or III quadrants.
II Quadrant
III Quadrant
o
o
o
θ =180 – 𝛼 =180 – 37° = 72 ,
θ = 180o + 𝛼 = 180o + 37o = 217o.
12. Determine the exact primary ratios for each of the following angles:
(Don't forget to consider and show the value of the ratio for the
reference angle)
[2K each]
a) sin 225°
Solution
sin 225° = sin (180°+ 45°) = – sin 45° = –
√2
.
2
b) cos 150°
Solution
cos 150° = cos (180° – 30°) = – cos 30° = –
√3
.
2
Communication [10 Marks]
13. Using the unit circle, explain why sin2 𝜃 + cos2 𝜃 = 1 is an identity.
[4C]
Answer:
Since r = 1, we get:
𝑦
sin 𝜃 = = y,
𝑟
𝑥
cos 𝜃 = = x.
𝑟
By Pythagorean Theorem,
x2 + y2 = r2.
(1)
Substitute x = cos 𝜃, y = sin 𝜃,
and r = 1 into (1):
sin2 𝜃 + cos2 𝜃 = 1.
14. Explain when an ambiguous case might exist. Include all necessary
conditions. As part of your explanation, provide examples (including a
diagram) of possible given value that would result in an ambiguous case.
You do not need to solve your own examples. [6C]
Answer
An ambiguous case exists when exist two
triangles ABC, such that ∠A = 𝜃, BC = a, and
AC = b. One of the triangles has an acute
∠ B, and the other has an obtuse ∠ B, which
is supplementary to the first.
An ambiguous case exists when h < a < b,
where h = b sin 𝜃.
For example, if 𝜃 = 30°, a = 4 cm and b = 6 cm, we have an ambiguous
case, since h = b sin = 3 cm and h < a < b. Using Sine Law, we can find
two possible values of ∠ B:
𝑎
𝑏
4
6
=
⟹
=
⟹ sin ∠B = 0.75
sin 𝜃
sin ∠𝐵
sin 30° sin ∠𝐵
∠B = sin– 1 (0.75) ≈ 48.6° or ∠B = 180° – 48.6° = 131.4°
Application [12 Marks]
15. Emma is on a 50 m high bridge and sees two boats in the water
below her. Emma estimates the angles of depression to be 38° for boat A
and 35° for boat B. How far apart are the two boats if the angle in
between Emma's lines of sight of the two boats is l10°. Draw a diagram
as part of your solution.
Solution
1) From ∆ AED:
𝐸𝐷
50
sin ∠EAD =
⟹ sin 38° =
AE =
50
sin 38°
𝐴𝐸
≈ 81.21 m.
2) From ∆ BED:
𝐸𝐷
sin ∠EBD =
⟹ sin 35° =
BE =
50
sin 35°
𝐴𝐸
𝐵𝐸
50
𝐵𝐸
≈ 87.17 m.
3) From ∆ ABE by Cosine Law:
AB2 = AE2 + BE2 – 2 AE ∙ BE cos ∠AEB
AB2 = 81.212 + 87.172 – 2(81.21) (87.17) cos 110°
AB ≈ 138.0 m.
16. A plane’s route is passing above Sinville and Tangland. The plane is
6.7 km directly away from Sinville and 5.3 km directly away from
Tangland. The angle of elevation from Sinville to the plane is 42°. How
far apart are Sinville and Tangland? Draw a diagram and consider
al1possibilities.
Solution
Determine the height
of the plane:
𝑃𝐻
sin ∠S =
⟹
𝑃𝑆
ℎ
sin 42°= ⟹
6.7
h = 6.7 sin 42°
h ≈ 4.48 km.
h < 5.3 < 6.7 ∴
2 triangles are
possible.
From ∆ SPT by Sine Law:
𝑃𝑆
𝑃𝑇
6.7
5.3
6.7 sin 42°
=
⟹
=
⟹ sin ∠PTS =
≈ 0.8459
sin ∠𝑃𝑇𝑆 sin ∠𝑆
sin ∠𝑃𝑇𝑆 sin 42°
5.3
∠PTS = sin– 1 (0.8459) ≈ 58° or ∠PTS = 180° – 58° =122°.
Case 1: ∠PT1S = 58o, ∠SPT1 = 180o – 42o – 58o = 80o.
From ∆ SPT1 by Sine Law:
𝑆𝑇1
sin ∠𝑆𝑃𝑇1
=
𝑃𝑇1
sin ∠𝑆
⟹
𝑆𝑇1
sin 800
=
5.3
sin 42o
⟹ ST1=
5.3 sin 80o
sin 420
= 7.8 km
Case 2: ∠PT2S = 122o, ∠SPT2 = 180o – 42o – 122o = 16o.
From ∆ SPT1 by Sine Law:
𝑆𝑇21
sin ∠𝑆𝑃𝑇2
=
𝑃𝑇2
sin ∠𝑆
⟹
𝑆𝑇2
sin 160
=
Answer: 2.2 km or 7.8 km.
5.3
sin 42o
⟹ ST2=
5.3 sin 16o
sin 420
= 2.2 km
Thinking
[12 Marks]
17. Prove the following identity. Use proper notation for full marks. [6T]
tan 𝜃 ∙ sec 𝜃 =
sin3 𝜃 + cos2 𝜃 sin 𝜃
1 − sin2 𝜃
.
Solution
R.S. =
=
sin3 𝜃 + cos2 𝜃 sin 𝜃
1 − sin2 𝜃
sin 𝜃
sin 𝜃
cos2 𝜃
=
cos 𝜃
∙
=
1
cos 𝜃
sin 𝜃(sin2 𝜃 + cos2 𝜃)
cos2 𝜃
= tan 𝜃 ∙ sec 𝜃 = L.S.
18. If tan2 𝜃 – 2 tan 𝜃 – 3 = 0, find all possible values of 𝜃, where
0o ≤ θ ≤ 360o. [6T]
Solution
tan2 𝜃 – 2 tan 𝜃 – 3 = 0, (tan 𝜃 3) (tan 𝜃 + 1) = 0,
tan 𝜃 – 3 = 0 or tan 𝜃 + 1= 0
tan 𝜃 = 3 or tan 𝜃 = –1
Case 1 tan 𝜃 = 3
Related angle 𝛼 = tan – 1 (3) ≈ 72°.
tan θ > 0 ∴ θ lies in the I or III quadrants.
I Quadrant
III Quadrant
θ = 𝛼 = 72o,
θ = 180o + 𝛼 = 180o + 72o = 252o.
Case 2 tan 𝜃 = –1
Related angle 𝛼 = tan – 1 (1) = 45°.
tan θ < 0 ∴ θ lies in the II or IV quadrants.
II Quadrant
IV Quadrant
o
o
o
o
θ = 180 – 𝛼 = 180 – 45 = 135 , θ = 360o – 𝛼 = 360o – 45o = 315o.
Answer: θ = 72o, 135o, 252o, 315o.