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Trigonometry Review
Name:
Multiple Choice Choose the correct answer.
1 The cosine ratio is defined as:
A
opposite
hypotenuse
B
adjacent
opposite
C
adjacent
hypotenuse
D
opposite
adjacent
cosine of an angle =
adjacent
hypotenuse
C
2 tan 3843, correct to four decimal places, is:
A 0.7934
B 0.7802
C 0.6962
D 0.8016
tan 3843= 0.8016
3 The correct length, to one decimal place, of the
side marked x is:
x = 15 sin 25°
D
= 6.3 m
B
A 13.6 m
B 6.3 m
C 7.0 m
D 6 cm
4 The size of the angle marked  is:
cos  =
4
6
 = 48°11´
D
A 41°49´
B 33°41´
C 56°19´
D 48°11´
New Signpost Mathematics Enhanced 9 5.1-5.3
Chapter Tests
–1–
© Pearson Australia 2009.
This page may be photocopied for classroom use.
5 A ladder leaning against a wall makes an angle
of 6312´ with the ground. If the ladder
reaches 1.95 m up the wall, the length of the
ladder is closest to:
sin 6312´ =
l=
1.95
l
1.95
sin 6312´
= 2.18 m
D
l
1.95 m
6312´
A 0.89 m
B 4.30 m
C 1.74 m
D 2.18 m
6 Which equation could be used to find the value
30.5
sin  =
of ?
79.3
30.5
73.2
79.3
73.2
79.3
B tan  =
73.2
30.5
C sin  =
73.2
79.3
D tan  =
30.5
79.3
7 The angle of elevation of the top of a 42 m tall
lookout tower from a landmark on the ground
is 34. How far is the landmark from the base
of the tower?
A 62.27 m
B 28.33 m
C 50.66 m
D 23.49 m
New Signpost Mathematics Enhanced 9 5.1-5.3
Chapter Tests
30.5
73.2
cos  =
73.2
79.3
A

A cos  =
tan  =
–2–
Let p be the distance of the landmark from the
base of the lookout tower.
tan 34 =
p=
42
p
42
tan 34
p = 62.27 m
A
© Pearson Australia 2009.
This page may be photocopied for classroom use.
8 A person standing on a cliff 258 m above sea
0.72 km = 720 m
level, sees a boat out at sea, 0.72 km from the
Let the angle of depression be .
base of the cliff. The angle of depression of the
boat from the person, to the nearest degree, is: tan  = 258
720

 = 19.7143
= 1943´
258 m
C
0.72 km
A 70
B 68
C 1943´
D 3
Short Answer
9 Find the value of the pronumeral in this
diagram:
cos 28° =
50
y
y = 50 ÷ cos 28°
= 56.63
10 Find the value of the pronumeral in each of the
following, correct to one decimal place.
(a)
(a) tan 5819´ =
a
16
16 tan 5819´ = a
a = 25.9 cm
16
58°19
(b) sin 4712´ =
a
15
s
s sin 4712´ = 15
s=
15
sin 4712´
s = 20.4 cm
(b)
47°12
15 cm
s
11 A wheelchair access ramp is 4.7 m long and it
makes an angle of 15°52´ with the ground.
How far above the lower end is the upper end
of the ramp?
New Signpost Mathematics Enhanced 9 5.1-5.3
Chapter Tests
–3–
sin 15°52´ = h ÷ 4.7,
so h = 4.7 sin 15°52´ = 1.3 m
© Pearson Australia 2009.
This page may be photocopied for classroom use.
12 Find the angle , to the nearest minute, in the
following diagrams.
(a) tan  =
23
18
 = 5157´
(a)
(b) cos  =

18 cm
17.3
26.4
 = 493´
23 cm
(b)
17.3 cm

26.4 cm
13 Stanley is standing 16 m from the base of a
skyscraper. How tall is the building if he has to
look through an angle of elevation of 758´ to
see the top? Assume Stanley’s height is 1.8 m.
tan 758´ = (h – 1.8)  16,
so h = 16 tan 758´ + 1.8 = 62.1 m
h
758´
16 m
1.8 m
14 Angela is standing at the top of a vertical cliff
5 m high. She looks through an angle of
depression of 27°22´ to look at a boat out to
sea. How far from the shore is the boat?
New Signpost Mathematics Enhanced 9 5.1-5.3
Chapter Tests
–4–
tan 62°38´ = d ÷ 5,
so d = 5 tan 62°38´ = 9.7 m
© Pearson Australia 2009.
This page may be photocopied for classroom use.