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P R AC T I C E T E S T, p a g e s 5 1 6 – 5 2 0
1. Multiple Choice The terminal arm of an angle u in standard
position is in Quadrant 4. Which statements are true?
I. tan u < 0 II. tan u > 0 III. sin u < cos u
IV. sin u > cos u
A. I and III
D. II and IV
B. I and IV
C. II and III
1
2. Multiple Choice An angle u is in standard position, with sin u = 5 .
Which statement could be true?
√
2 6
B. cos u = 5
4
A. cos u = 5
√
C. cos u =
26
5
2
D. cos u = 5
3. Point P(-2, 7) is on the terminal arm of an angle u in standard
position.
a) Determine sin u, cos u, and tan u.
The distance
√ of P from the origin is r.
Use: r x2 y2
Substitute: x 2, y 7
r
r
√
( 2)2 72
√
53
y
sin U r
y
x
cos U r
tan U x
2
7
√
√
53
53
7
, or 3.5
2
b) Determine the measure of u to the nearest degree.
7
Use: sin U √
53
The reference angle is:
sin1 a√ b ⬟ 74°
7
53
U is approximately 180° 74°, or 106°.
4. Without using a calculator, determine the exact value of each
expression.
a) (sin 45°)(cos 45°)
b)
1
Substitute: sin 45° √ ,
2
1
cos 45° √
2
(sin 45°)(cos 45°)
tan 45°
cos 60°
Substitute: tan 45° 1,
cos 60° 0.5
1
tan 45°
cos 60°
0.5
2
a√ b a√ b
1
2
1
2
1
2
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c) (tan 120°)(tan 210°)
√
Substitute: tan 120° 3,
1
tan 210° √
3
(tan 120°)(tan 210°)
√
( 3 ) a√ b
1
3
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d) sin 330° + cos 120°
Substitute: sin 330° 0.5,
cos 120° 0.5
sin 330° cos 120°
0.5 (0.5)
1
1
5. For which angles in standard position are the following statements true?
Give the angle measures to the nearest degree for 0° ≤ u ≤ 360°.
1
2
a) sin u = - √
The reference angle is:
sin
1
1
a√ b 45°
2
sin U is negative in
Quadrants 3 and 4, so
U 180° 45°, or 225°
or
U 360° 45°, or 315°
3
b) cos u = 4
The reference angle is:
cos1 a b ⬟ 41°
3
4
cos U is positive
in Quadrants 1 and 4, so
U ⬟ 41°
or
U ⬟ 360° 41°, or 319°
6. Solve each triangle. Give the angle measures to the nearest degree
and the side lengths to the nearest tenth of a unit.
a) In ⌬ABC, AB = 20 m, ∠ A = 65°, and ∠ B = 40°
Sketch a diagram. Since 2 angles and the contained side
are given, only one triangle can be drawn.
∠C 180° (40° 65°)
75°
20 m
a
c
Use:
sin A
B
40⬚
sin C
Substitute: ∠ A 65°
A
C
20
b
sin B
sin 75°
∠ C 75°, c 20
Use:
a
20
sin 65°
sin 75°
20 sin 65°
a
sin 75°
Substitute: ∠B 40°
b
65⬚
20 sin 40°
sin 75°
a 18.7655. . .
b 13.3092. . .
So, BC ⬟ 18.8 m and AC ⬟ 13.3 m
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b) In ⌬KMN, KM = 25.0 m, MN = 24.6 m, and ∠ K = 75°
Check for the ambiguous case.
The ratio of the side opposite ∠ K to the side adjacent to ∠ K is:
MN
24.6
, which is 0.984
KM
25.0
sin 75° 0.9659. . .
Since sin 75°<0.984<1, then there is an ambiguous case, and
2 triangles can be constructed: ≤ KMN1 is acute; ≤ KMN2 is obtuse.
Sketch a diagram.
M
24.6 m
25.0 m
75⬚
K
N2
N1
This diagram is not drawn to scale.
In ≤ KMN1
Determine ∠ N1.
Use:
sin N1
sin K
n k
Substitute: ∠ K 75°,
n 25, k 24.6
sin N1
sin 75°
25
24.6
25 sin 75°
sin N1 24.6
∠N1 sin1 a
25 sin 75°
b
24.6
∠ N1 79.0014. . .°
So, ∠M 180° (75° 79.0014. . .°)
25.9985. . . °
Use:
m
k
sin M
sin K
Substitute: ∠ M 25.9985. . .°
∠ K 75°, k 24.6
In ≤ KMN2
∠ N2 180° ∠N1
100.9985. . .°
So, ∠ M 180° (75° 100.9985. . .°)
4.0014. . .°
Use:
m
k
sin M
sin K
Substitute: ∠M 4.0014. . .°,
∠K 75°, k 24.6
m
24.6
sin 4.0014. . .°
sin 75°
m
24.6 sin 4.0014. . .°
sin 75°
m 1.7771. . .
So, ∠N ⬟ 101°, ∠M ⬟ 4°,
and KN ⬟ 1.8 m
m
24.6
sin 25.9985. . .°
sin 75°
24.6 sin 25.9985. . .°
m
sin 75°
m 11.1637. . .
So, ∠ N ⬟ 79°, ∠ M ⬟ 26°,
and KN ⬟ 11.2 m
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c) In ⌬PQR, PQ = 15.0 cm, PR = 11.0 cm, and QR = 10.5 cm
Sketch a diagram. Since 3 sides are given,
only one triangle can be drawn.
Use: r2 p2 q2 2pq cos R
Substitute: r 15, p 10.5, q 11
P
15.0 cm
15 10.5 11 2(10.5)(11) cos R
231 cos R 6.25
2
2
11.0 cm
2
Q
10.5 cm
6.25
cos R 231
R
∠ R 88.4496. . .°
Use: q2 p2 r2 2pr cos Q
Substitute the values above.
112 10.52 152 2(10.5)(15) cos Q
315 cos Q 214.25
cos Q 214.25
315
∠ Q 47.1439. . .°
∠P 180° (88.4496. . .° 47.1439. . .°)
44.4064. . .°
So, ∠P ⬟ 44°, ∠ Q ⬟ 47°, ∠ R ⬟ 88°
7. A cross-country runner runs due east for 6 km, then changes course
to E25°N and runs another 9 km. To the nearest tenth of a
kilometre, how far is the runner from her starting point?
Sketch a diagram.
In ≤ STV,
∠T 180° 25°
155°
To determine SV, use:
t2 s2 v2 2sv cos T
Substitute: s 9, v 6, ∠ T 155°
t2 92 62 2(9)(6) cos 155°
N
V
N
9 km
25⬚
S
6 km
T
E
√
t 92 62 2(9)(6) cos 155°
t 14.6588. . .
The runner is approximately 14.7 km from her starting point.
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8. In parallelogram BCDE, BC = 10 cm, CD = 15 cm, and ∠ B = 135°;
determine the lengths of the diagonals to the nearest tenth of a
centimetre.
Sketch a diagram.
∠C 180° 135°
45°
In ≤ BCD, to determine BD, use:
c2 b2 d2 2bd cos C
Substitute: b 15, d 10, ∠ C 45°
c2 152 102 2(15)(10) cos 45°
15 cm
C
10 cm
135⬚
B
D
E
√
c 152 102 2(15)(10) cos 45°
c 10.6239. . .
Diagonal BD is approximately 10.6 cm long.
In ≤ BCE, to determine CE, use:
b2 c2 e2 2ce cos B
Substitute: c 15, e 10, ∠B 135°
b2 152 102 2(15)(10) cos 135°
√
b 152 102 2(15)(10) cos 135°
b 23.1761. . .
Diagonal CE is approximately 23.2 cm long.
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