MCR 3U Grade 11 University
TRIGONOMETRY CHAPTER TEST
1. For each point,
 sketch the angle in standard position;
 determine all six trigonometric ratios;
 Determine, where appropriate, the value of the related acute angle,
to the nearest degree;
 Determine the value of the principal angle, to the nearest degree.
a) P (– 3, 0)
Solution
Principal angle 𝜽 = 180o.
sin 180o = 0
cos 180o = – 1
tan 180o = 0
cot 180o = DNE
csc 180o = DNE
sec 180o = – 1
b) P (– 4, – 3)
Solution
x = – 4, y = – 3,
𝟐
𝟐
r = √(– 𝟒) + (– 𝟑) = 5
𝒚
sin 𝜽 = = –
𝒓
𝒚
tan 𝜽 = =
𝒙
𝒓
𝟑
𝟓
𝒙
cos 𝜽 = = –
𝒓
𝒙
cot 𝜽 = =
𝟒
csc 𝜽 = = –
𝒚
𝟑
𝟓
𝟑
𝒚
𝒓
𝟒
𝟑
𝟓
𝟑
sec 𝜽 = = –
𝒙
𝟒
𝟓
𝟒
Related ange 𝜶 = sin– 1 ( ) ≈ 37o.
𝟓
o
Principal angle 𝜽 ≈ 180 +37o = 117o.
2. Given an angle 𝜽 where 0o ≤ 𝜽 ≤ 360o, determine all possible values
for 𝜽.
𝟏
a) sin 𝜽 = –
𝟐
Solution
𝟏
Related ange 𝜶 = sin– 1 ( ) = 30o.
𝟐
Sine sin 𝜽 is negative in the third and forth quadrants, we get:
𝜽 = 180o + 𝜶 = 180o + 30o = 210o or 𝜽 = 360o – 𝜶 = 360o – 30o = 330o.
Answer: 210o or 330o.
b) cos 𝜽 =
Solution
√𝟑
𝟐
√𝟑
Related ange 𝜶 = cos– 1 ( ) = 30o.
𝟐
Sine cos 𝜽 is positive in the first and forth quadrants, we get:
𝜽 = 𝜶 = 30o or 𝜽 = 360o – 𝜶 = 360o – 30o = 330o.
Answer: 30o or 330o.
c) cot 𝜽 = – 1
Solution
𝟏
tan 𝜽 =
=–1
𝐜𝐨𝐭 𝜽
Related ange 𝜶 = tan– 1 (1) = 45o.
Sine tan 𝜽 is negative in the second and forth quadrants, we get:
𝜽 = 180o – 𝜶 =180o – 45o = 135o or 𝜽 = 360o – 𝜶 = 360o – 45o = 315o.
Answer: 135o or 315o.
d) sec 𝜽 = – 2
Solution
𝟏
𝟏
cos 𝜽 =
=–
𝐬𝐞𝐜 𝜽
𝟐
𝟏
Related ange 𝜶 = cos– 1 ( ) = 60o.
𝟐
Sine cos 𝜽 is negative in the second and third quadrants, we get:
𝜽 = 180o – 𝜶 =180o – 60o = 120o or 𝜽 = 180o + 𝜶 = 180o + 60o = 240o.
Answer: 120o or 240o.
𝟓
3. Given cos 𝜽 = − , where the terminal arm of angle 𝜽 lies in the third
𝟏𝟑
quadrant, evaluate each trigonometric expression.
a) sin 𝜽 ∙ cos 𝜽
Solution
x = – 5, r = 13
The angle 𝜽 lies in the third quadrant ∴ y < 0.
x2 + y2 = r2 ∴ y2 = r2 – x2 = 132 – (– 5)2
= 169 – 25 = 144 ∴ y = – 12.
𝒚
𝟏𝟐
sin 𝜽 = = − ,
𝒓
𝟏𝟑
sin 𝜽 ∙ cos 𝜽 = (−
𝟏𝟐
𝟏𝟑
) (−
b) tan 𝜽 ∙ cot 𝜽
Solution
tan 𝜽 ∙ cot 𝜽 = tan 𝜽 ∙
𝟓
𝟏𝟑
𝟏
𝐭𝐚𝐧 𝜽
)=
𝟔𝟎
𝟏𝟔𝟗
.
= 1.
4. Prove each identity. Use different methods for parts a) and b). State
any restrictions on the variables. Explain why these identities are called
Pythagorean identities.
a) tan2 𝜽 + 1 = sec2 𝜽
Solution
𝒚
𝒓
tan 𝜽 = , sec 𝜽 =
𝒙
tan2 𝜽 + 1 =
𝒙
𝒚𝟐
𝒚𝟐 + 𝒙𝟐
𝒙
𝒙𝟐
𝟐 + 1 =
=
𝒓𝟐
𝒓 𝟐
𝒙
𝒙
2
𝟐 = ( ) = sec 𝜽
Let ABC be a right triangle with AC = 1
and ∠A = 𝜽. Then
BC = AC tan ∠A = tan 𝜽,
𝑨𝑪
𝟏
AB =
=
= sec 𝜽.
𝐜𝐨𝐬 ∠𝑨 𝐜𝐨𝐬 𝜽
By Pythagoreans Theorem,
AC2 + BC2 = AB2 ⟹ 1 + tan2 𝜽 = sec2 𝜽
b) cot2 𝜽 + 1 = csc2𝜽
Solution
2
cot 𝜽 + 1 =
𝐜𝐨𝐬𝟐 𝜽
𝐬𝐢𝐧𝟐 𝜽
+1=
𝐜𝐨𝐬𝟐 𝜽 + 𝐬𝐢𝐧𝟐 𝜽
𝐬𝐢𝐧𝟐 𝜽
=
𝟏
𝐬𝐢𝐧𝟐 𝜽
= csc2 𝜽.
5. The sides of a triangle are 16 cm, 19 cm, and 21 cm. Determine its
angles to the nearest degree.
Solution
Let a = 16, b = 19, c = 21,
cos A =
𝒃𝟐 + 𝒄𝟐 − 𝒂𝟐
𝟐𝒃𝒄
=
𝟏𝟗𝟐 + 𝟐𝟏𝟐 − 𝟏𝟔𝟐
𝟐(𝟏𝟗)(𝟐𝟏)
=
𝟑𝟔𝟏 + 𝟒𝟒𝟏− 𝟐𝟓𝟔
𝟓𝟒𝟔
=
𝟕𝟗𝟖
= 0.6842
𝟕𝟗𝟖
∠A = cos – 1 (0.6842) ≈ 47o.
cos B =
𝒂𝟐 + 𝒄𝟐 − 𝒃𝟐
𝟐𝒂𝒄
=
𝟏𝟔𝟐 + 𝟐𝟏𝟐 − 𝟏𝟗𝟐
𝟐(𝟏𝟔)(𝟐𝟏)
=
𝟐𝟓𝟔 + 𝟒𝟒𝟏 − 𝟑𝟔𝟏
𝟔𝟕𝟐
=
𝟑𝟑𝟔
𝟔𝟕𝟐
= 0.5
∠B = cos – 1 (0.5) = 60o.
cos C =
𝒂𝟐 + 𝒃𝟐 − 𝒄𝟐
𝟐𝒂𝒃
=
𝟏𝟔𝟐 + 𝟏𝟗𝟐 − 𝟐𝟏𝟐
𝟐(𝟏𝟔)(𝟏𝟗)
∠C = cos – 1 (0.2895) ≈ 73o.
Answer: 47o, 60o, 73o.
=
𝟐𝟓𝟔+ 𝟑𝟔𝟏− 𝟒𝟒𝟏
𝟔𝟎𝟖
=
𝟏𝟕𝟔
𝟔𝟎𝟖
= 0.2895
6. For each triangle, calculate the value of w to the nearest tenth of a
metre.
a)
Solution
From ∆ ABD: tan 49o =
BD =
𝑨𝑫
≈
𝐭𝐚𝐧 𝟒𝟗𝐨
𝟔𝟎
𝟏.𝟏𝟓
𝑨𝑫
≈
𝐭𝐚𝐧 𝟓𝟑𝐨
𝟔𝟎
𝟏.𝟑𝟑
𝑩𝑫
⟹
≈ 52.2 m
From ∆ ACD: tan 53o =
CD =
𝑨𝑫
𝑨𝑫
𝑪𝑫
⟹
≈ 47.2 m
w = BC = BD + CD = 52.2 + 47.2 = 99.4
b)
Solution
∠GEH = 48o – 40o = 8o.
From ∆ GEH by Sine Law:
𝑮𝑯
𝒐 =
𝐬𝐢𝐧 𝟖
w=
𝑬𝑯
𝐨 ⟹
𝐬𝐢𝐧 𝟒𝟎
𝟓.𝟓 𝐬𝐢𝐧 𝟖𝒐
𝐬𝐢𝐧 𝟒𝟎𝐨
𝒘
𝒐 =
𝐬𝐢𝐧 𝟖
≈ 1.2 m
𝟓.𝟓
𝐬𝐢𝐧 𝟒𝟎𝐨
7. Given each set of information, determine how many triangles can be
drawn. Then, calculate all unknown side lengths to the nearest tenth and
all unknown interior angles to the nearest degree.
a) a = 1.5 cm, b = 2.8 cm, and ∠A = 41o.
Solution
h = 2.8 sin 41o ≈ 1.8 cm
a = 1.5 < h ∴ 0 triangles
b) a = 2.1 cm, c = 6.1 cm, and ∠A = 20o
Solution
h = 6.1 sin 20o ≈ 2.1 cm
a = 2.1 = h ∴ 1 triangle
AC = 6.1 cos 20o = 5.7 cm
8. To estimate the amount of usable lumber in a tree, Luca must first
estimate the height of the tree. From points A and B on the ground, he
determined that the angles of elevation for a certain tree were 41 o and
52o, respectively. The angle formed at the base of the tree between the
points A and B is 90o, and AB = 10 m. If the tree is perpendicular to the
ground, what is its height to the nearest metre?
Solution
Let CD = h.
1) ∠ADC = 90o – 41o = 49o
From ∆ACD: AC = h tan 49o
2) ∠BDC = 90o – 52o = 38o
From ∆BCD: BC = h tan 38o
3) From ∆ABC by Pythagorean Theorem:
AC2 + BC2 = AB2 ⟹
(ℎ tan 49o )2 + (ℎ tan 38o )2 = (10)2
h2 (tan2 49o + tan2 38o) = 100
10
h=
≈7m
2
o
2
o
√tan 49
+ tan 38
9. In a triangle ABC, ∠A = 45o, ∠B = 60o, BC = 6 cm. Solve the triangle.
For the sides of the triangle, give your answers in exact radical form.
Solution
∠C = 180o – 45o – 60o = 75o.
By Sine Law:
𝑨𝑪
𝑩𝑪
𝟔 𝐬𝐢𝐧 𝟔𝟎°
=
⟹ AC =
𝐬𝐢𝐧 𝟔𝟎°
𝐬𝐢𝐧 𝟒𝟓°
𝐬𝐢𝐧 𝟒𝟓°
√𝟑
=
𝟔( 𝟐 )
𝟏
)
√𝟐
(
= 3√𝟔.
Let AB = x. By Cosine Law:
AB2 + BC2 – 2 AB ∙ BC cos B = AC2 ⟹
x2 + 62 – 2 x ∙ 6 cos 60o = (3√𝟔)2
x2 + 36 – 6x = 54
x2 – 6x – 18 = 0
x=
𝟔 ± √(−𝟔)𝟐 − 𝟒(𝟏)(−𝟏𝟖)
𝟐(𝟏)
=
𝟔 ± √𝟑𝟔+ 𝟕𝟐
𝟐
=
𝟔 ± √𝟑𝟔 ∙ 𝟑
𝟐
x > 0 ⟹ AB = x = 3 + 3√𝟑.
Answer: ∠C = 75o, AC = 3√𝟔, AB = 3 + 3√𝟑.
=
𝟔 ± 𝟔√ 𝟑
𝟐
= 3 ± 3√𝟑