MCR3U Grade 11 University
Trigonometry 11 Test 1
Thinking, Inquiry & Problem Solving
Knowledge & Understanding
Application
Communication
/ 11
/ 20
/9
/6
Weighted Mark: ________%
Knowledge & Understanding
1. Find the measure of the angle A in a triangle ABC if AB = 4.8 cm, BC
= 4.1 cm, and AC = 6.7 cm. Round your answer to the nearest degree.
[K 2]
Solution
a = BC = 4.1, b = AC = 6.7, c = AB = 4.8,
cos A =
𝑏2 + 𝑐 2 − 𝑎2
2𝑏𝑐
=
6.72 + 4.82 − 4.12
2(6.7)(4.8)
=
44.89 + 23.04 – 16.81
64.32
=
51.12
64.32
≈ 0.7948
∠A = cos – 1 (0.7948) ≈ 37o.
cos B =
𝑎2 + 𝑐 2 − 𝑏2
2𝑎𝑐
=
4.12 + 4.82 − 6.72
2(4.1)(4.8)
=
16.81 + 23.04 − 44.89
39.36
=
−5.04
39.36
≈ – 0.128
∠B = cos – 1 (– 0.128) ≈ 97o.
cos C =
𝑎2 + 𝑏2 − 𝑐 2
2𝑎𝑏
=
4.12 + 6.72 − 4.82
2(4.1)(6.7)
=
16.81+ 44.89 − 23.04
54.94
=
38.66
54.94
≈ 0.7037
∠C = cos – 1 (0.7037) ≈ 45o.
Answer: 37o, 97o, 45o.
Note: The sum of the angles adds up to 179o, not 180o, because of the
rounding.
7
2. Given: tan θ =− , 270o < θ < 360o.
24
a) State the x- and y-coordinates of a point P(x, y) on the terminal arm of
the angle and its distant r = OP from the origin: [K 2]
x = _____
y = _____
r = ______
Solution
7
𝑦
tan θ =− = ⟹ x = – 24, y = 7 or x = 24, y = –7
24
𝑥
o
270 < θ < 360o ⟹ x > 0, y < 0 ⟹ y = – 7, x = 24 ⟹ P (24, – 7).
r = √𝑥 2 + 𝑦 2 = √(24)2 + (−7)2 = 25.
b) State the other five trigonometric ratios of the angle θ: [K 5]
sin θ = _____ cos θ = _____ cot θ = _____ sec θ = _____ csc θ = _____
Solution
𝑦
7
𝑥 24
𝑥
24
sin θ = = − ,
cos θ = = ,
cot θ = = − ,
𝑟
𝑟
25
25
𝑥
24
sec θ = =
,
𝑟
𝑟
25
25
𝑦
7
𝑦
7
csc θ = = − ,
c) Determine the value of θ to the nearest degree: θ = _____ [K 2]
Solution
7
Determine the related angle: 𝛼 = sin – 1 ( ) ≈ 16o
25
3. Find the exact value for each of the following:
a) cot 330o = _____ [K 2]
Solution
cot 330o = cot (360o – 30o) = – cot 30o = – √3
b) sin 225o = _____ [K 2]
Solution
sin 225o = sin (180o + 45o) = – sin 45o = –
√2
2
4. The point P (– 5, – 7) is on the terminal arm of an angle in standard
position. [K 3]
a) Sketch the principal angle.
Solution
b) Determine the measure of the principal angle to the nearest tenth of
degree.
Solution
7
Determine the related angle: 𝛼 = tan – 1 ( ) ≈ 54.5o
5
o
Determine the principal angle: 𝜗 = 180 + 54.5o = 234.5o
Application
5. Solve Δ ABC if AB = 11 cm, AC = 9 cm, and ∠B = 48o. [A 3]
Solution
ℎ
From ∆ABH: sin 48o = ⟹
11
h = 11 sin 48o ≈ 8.2 cm
Since b = AC = 9 cm, we have:
h < b < c ∴ 2 triangles.
From ∆ABC by Sine Law:
𝐴𝐵
𝐴𝐶
11
9
=
⟹
=
o ⟹
sin 𝐶
sin 𝐵
sin 𝐶
11 sin 48o
sin 48
sin C =
= 0.9083.
9
C = sin– 1 (0.9083) ≈ 65o or C = 180o – 65o = 115o.
Case 1: ∠A𝐶1 B = 65o, ∠BAC1 = 180o – 48o – 65o = 67o
From ∆ABC1 by Sine Law:
𝐴𝐶1
sin 𝐵
=
𝐵𝐶1
sin ∠𝐵𝐴𝐶1
⟹
9
sin 48
0 =
𝐵𝐶1
sin 67
o ⟹ 𝐵𝐶1 =
9 sin 67o
sin 480
=11.15 cm
Case 2: ∠A𝐶2 B = 115o, ∠BAC2 = 180o – 48o – 115o = 17o
From ∆ABC2 by Sine Law:
𝐴𝐶2
sin 𝐵
=
𝐵𝐶2
sin ∠𝐵𝐴𝐶2
⟹
9
sin 48
0 =
𝐵𝐶2
sin 17
o ⟹ 𝐵𝐶2 =
9 sin 17o
sin 480
=3.54 cm
6. Solve for θ: sin θ = – 0.5, 0o < θ < 360o. [A 3]
Solution
Determine the related angle: 𝛼 = sin – 1 0.5 = 30o.
Since sin θ < 0, θ is in the III or in the IV quadrant. Therefore,
θ = 180o + 𝛼 = 210o or θ = 360o – 𝛼 = 330o.
7. Solve for θ: sec θ = – 2.6, 0o < θ < 360o. [A 3]
Solution
1
10
5
sec θ = – 2.6 ⟹ cos θ = – = – = –
2.6
26
13
–1 5
( )
13
Determine the related angle: 𝛼 = sin
= 22o.
Since sin θ < 0, θ is in the III or in the IV quadrant. Therefore,
θ = 180o + 𝛼 = 210o or θ = 360o – 𝛼 = 330o.
Thinking, Inquiry & Problem Solving
8. a) Using exact values, verify the following identity for θ = 60o: [T 2]
1 + cot2 θ = csc2 θ.
Solution
1
2
1
4
3
3
L.S. = 1 + cot2 60o = 1 + ( ) = 1 + =
2
2
R.S. = csc2 60o = ( ) =
√3
√3
4
3
∴ L.S. = R.S.
b) State the restrictions for θ, 0o < θ < 360o. [T 2]
Solution
sin θ ≠ 0 ∴ θ ≠ 0o, 180o, 360o
c) Prove the identity for all values of θ which satisfy the restrictions.
[T 2]
Solution
cos2 𝜃
sin2 𝜃+ cos2 𝜃
1
L.S. = 1 + cot θ = 1 + 2 =
= 2 = csc2 θ.
sin 𝜃
sin2 𝜃
sin 𝜃
o
sin θ ≠ 0 ∴ θ ≠180 n, where n is an integer.
2
9. Prove the identity: (cot x) (cos x) + sin x = csc x. [T 2]
Solution
cos 𝑥
(cot x) (cos x) + sin x =
∙ cos x + sin x
=
sin 𝑥
cos2 𝑥 + sin2 𝑥
sin 𝑥
=
1
sin 𝑥
= csc x.
10. A crow sitting on the top of a tree with a piece of cheese in her beak
sees a fox and a wolf in the same line of sight, the fox is 10 m ahead of
the wolf. The crow is 70 m from the fox and observes the wolf at the
angle of depression of 16o. Determine the angle of depression the crow
observes the fox and the height of the tree. [T 3]
Solution
From ∆WCF by Sine Law:
sin ∠𝑊𝐶𝐹
𝑊𝐹
=
sin ∠𝐶𝑊𝐹
𝐶𝐹
1
⟹
sin ∠𝑊𝐶𝐹
10
=
sin 16o
70
⟹
sin ∠𝑊𝐶𝐹= sin 16o ≈ 0.0394 ⟹ ∠WCF ≈ sin– 1 (0.0394) ≈ 2o
7
∠CFB =∠CWF + ∠WCF ≈ 16o + 2o = 18o (Exterior Angle Theorem)
From ∆CFB: CB = CF sin ∠CFB = 70 sin 18o ≈ 21.6 m
Answer: 18o, 21.6 m
Communication
11. In a triangle ABC, ∠A = 75o, AB = 5.8 m, BC = 8.7 m. Without
finding the value of the angle C, determine the number of possible
values of this angle. [C 2]
Solution
Since BC > AB, we can construct 1 triangle with the given data,
therefore, the number of possible values of angle C equals 1.
12. For each of the following questions state whether the statement is
true or false. Justify your answer.
a) sec2 θ + csc2 θ = sec2 θ ∙ csc2 θ; [C 2]
Solution
The statement is true:
L.S. = sec θ + csc θ =
2
=
2
1
cos 2 𝜃
∙
sin2 𝜃
=
1
cos 2 𝜃
1
cos 2 𝜃
∙
+
1
sin2 𝜃
1
sin2 𝜃
=
sin2 𝜃 + cos 2 𝜃
cos 2 𝜃 ∙ sin2 𝜃
= sec2 θ ∙ csc2 θ = R.S.
b) tan 270o ∙ cot 270o = 1. [C 2]
Solution
This statement is false, since tan 270o is undefined:
sin 270o
−1
tan 270o =
= – DNE
sin 270o
0
The identity tan 𝜃 ∙ cot 𝜃= 1 is true for 𝜃 ≠ 90on, where n is an integer.