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9
NON-FOUNDATION
More about Trigonometry (II)
Name :
9A
9.1
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Date :
Mark :
Area of a Triangle
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Key Concepts and Formulae
A
Area of a triangle
Area of ABC =
1
1
1
ab sin C = bc sin A = ca sin B
2
2
2
c
B
b
C
a
(Unless otherwise specified, numerical answers should either be exact or correct to 3 significant
figures.)
1.
Find the areas of the following triangles.
(a)
(b)
A
A
70o
5 cm
7.5 cm
130o
B
B
12 cm
C
C
Solution
(a) Area of the triangle
1
= ⎡⎢ × ( 5 ) × ( 7.5 ) × sin ( 70°
2
⎣
= ( 17.6 cm2 ) (cor. to 3 sig. fig.)
92
(b) Area of the triangle
)⎤⎥ cm 2
⎦
=
⎛1
⎞
× 12 × 12 × sin 130° cm2
⎝2
⎠
= 55.2 cm2 (cor. to 3 sig. fig.)
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9
2.
More about Trigonometry (II)
A
Find the value of x in ABC.
118o
8 cm
Solution
x cm
2
area = 23 cm
1
× AB × AC × sin A = area of ABC
2
B
C
1
(8)(x ) sin 118° = 23
2
x = 6.51 (cor. to 3 sig. fig.)
3.
In PQR, θ is an acute angle. Find θ.
R
Solution
15 cm
1
× PR × PQ × sin θ = area of PQR
2
1
(15)(8) sin θ = 30
2
P
area = 30 cm2
θ
8 cm
Q
sin θ = 0.5
θ = 30°
4.
The area of ABC is 80 cm2. If AB = 13 cm , AC = 15 cm and ∠BAC is an obtuse angle,
find ∠BAC.
Solution
Let θ be ∠BAC.
1
(13)(15) sin θ = 80
2
sin θ =
32
39
θ = 55.1° (cor. to 3 sig. fig.) or θ = 125° (cor. to 3 sig. fig.)
Q θ is an obtuse angle
∴
∠BAC = 125°
93
Measures, Shape and Space
5.
D
Find the area of parallelogram ABCD.
Solution
C
11 cm
AB = CD
(properties of parallelogram)
AD = CB
(properties of parallelogram)
∠DAB = ∠DCB
(properties of parallelogram)
70o
A
B
16 cm
∴ Area of parallelogram ABCD = 2 × area of ABD
⎡1
⎤
= 2⎢ (11)(16) sin 70°⎥ cm2
⎣2
⎦
= 165 cm2 (cor. to 3 sig. fig.)
6.
In parallelogram ABCD, AD = 12 cm , ∠ABC = 130°
and its area is 82 cm2. Find DC.
Solution
12 cm
D
130o
B
Let DC = x cm .
∠ADC = ∠ABC = 130°
(properties of parallelogram)
Area of parallelogram ABCD = 2 × area of ACD
⎡1
⎤
82 = 2 × ⎢ (x )(12) sin 130°⎥
⎣2
⎦
x = 8.92 (cor. to 3 sig. fig.)
∴ DC = 8.92 cm
94
A
C
9
7.
More about Trigonometry (II)
In the figure, AD = 5.5 cm , AC = 7 cm , AB = 10 cm ,
∠DAC = 30° and ∠CAB = 60° . Find the area of quadrilateral
ABCD.
5.5 cm
30o
A
D
60o
7 cm
C
Solution
10 cm
Area of quadrilateral ABCD
= area of ACD + area of ABC
=
1
⎛1
⎞
× 7 × 5.5 × sin 30° + × 7 × 10 × sin 60° cm2
⎝2
⎠
2
B
= 39.9 cm2 (cor. to 3 sig. fig.)
8.
In the figure, O is the centre of the circle. Find the area of the shaded
segment.
Solution
∠AOB = 180° − 25° − 25°
O
15 cm
25o
A
B
= 130°
Area of the shaded segment
= area of sector OAB − area of OAB
1
130°
⎛
⎞
= π × 152 ×
−
× 15 × 15 × sin 130° cm2
⎝
⎠
2
360°
= 169 cm2 (cor. to 3 sig. fig.)
95
Measures, Shape and Space
9.
Find the area of a rhombus with side 15 cm and an acute angle of 65°.
Solution
1
Area of the rhombus = ⎛ 2 × × 15 × 15 × sin 65°⎞ cm2
⎝
⎠
2
= 204 cm2 (cor. to 3 sig. fig.)
10. In ABD, C is a point on BD such that CD = 3.8 cm , AC = 9.5 cm .
Find the area of ABD.
D
3.8 cm
Solution
C
∠ACD = 30° + 70°
(ext. ∠ of )
9.5 cm
= 100°
Area of ABD = area of ABC + area of ACD
=
1
⎛1
⎞
× 10 × 9.5 × sin 30° + × 9.5 × 3.8 × sin 100° cm2
⎝2
⎠
2
= 41.5 cm2
96
A
70o
30o
10 cm
B
9
NON-FOUNDATION
More about Trigonometry (II)
Name :
9B
9.2
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Date :
Mark :
The Sine Formula
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Key Concepts and Formulae
A
The sine formula
a
b
c
=
=
sin A
sin B
sin C
c
b
B
C
a
(In this exercise, unless otherwise specified, numerical answers should either be exact or correct to
3 significant figures.)
Find the unknowns in the following triangles. (1–6)
C
1.
C
2.
60o
8 cm
y cm
x cm
y cm
A
70o
A
35o
x cm
120o
B
B
5 cm
Solution
Solution
B = 180° − ( 60° ) − ( 70° )
= ( 50° )
C = 180° − 35° − 120°
By the sine formula,
By the sine formula,
x
sin ( 50° )
=
( 5 )
sin ( 60° )
x = ( 4.42
y
sin ( 70° )
=
)(cor. to 3 sig. fig.)
( 5 )
sin ( 60° )
y = ( 5.43
)(cor. to 3 sig. fig.)
= 25°
x
8
=
sin 25°
sin 120°
x = 3.90 (cor. to 3 sig. fig.)
y
8
=
sin 35°
sin 120°
y = 5.30 (cor. to 3 sig. fig.)
97
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Measures, Shape and Space
3.
D
4.
C
67o
y cm
9 cm
o
35o 40
a cm
d cm
b cm
45o
50o
A
B
x cm
Solution
A
B
c cm
C
Solution
B = 180° − 50° − 67°
= 63°
Consider ABD.
∠DBA = 180° − 45° − 35°
By the sine formula,
x
9
=
sin 67°
sin 63°
∴
5 cm
= 100°
By the sine formula,
x = 9.30 (cor. to 3 sig. fig.)
y
9
=
sin 50°
sin 63°
y = 7.74 (cor. to 3 sig. fig.)
a
5
=
sin 100°
sin 35°
∴
a = 8.58 (cor. to 3 sig. fig.)
b
5
=
sin 45°
sin 35°
b = 6.16 (cor. to 3 sig. fig.)
C = 180° − (35° + 40°) − 45°
= 60°
Consider BCD.
By the sine formula,
c
b
=
sin 40°
sin 60°
c =
5 sin 45°
sin 40°
×
sin 35°
sin 60°
= 4.58 (cor. to 3 sig. fig.)
d
b
=
sin (45° + 35°)
sin 60°
d =
5 sin 45°
sin 80°
×
sin 35°
sin 60°
= 7.01 (cor. to 3 sig. fig.)
98
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5.
6.
D
18 cm
D
o
130
7 cm
A
More about Trigonometry (II)
35o
C
o
80
7 cm
70o
30o
u cm
B
v cm
C
A
x cm
B
Solution
Solution
DB = DC = 7 cm
∠DAC = 180° − 130° − 35° (∠ sum of )
(base ∠s, isos. )
∠DBC = 70°
∠DBA = 180° − 70° (adj. ∠s on st. line)
= 15°
By the sine formula,
= 110°
AC
DC
=
sin ∠ADC
sin ∠DAC
∠ADB = 180° − 30° − 110° (∠ sum of )
= 40°
∴
By the sine formula,
7
u =
sin 30°
18 sin 130°
cm
sin 15°
∠CBA
u cm
DB
=
sin ∠ADB
sin ∠DAB
∴
AC =
= 180° − 35° − 80° (int. ∠s, AB // DC)
= 65°
× sin 40°
= 9.00 (cor. to 3 sig. fig.)
∠BDC = 180° − 70° − 70°
= 40°
By the sine formula,
AB
AC
=
sin ∠ACB
sin ∠CBA
∴
By the sine formula,
x =
18 sin 130°
sin 80°
×
sin 15°
sin 65°
= 57.9 (cor. to 3 sig. fig.)
v cm
DC
=
sin ∠BDC
sin ∠DBC
v =
7
sin 70°
× sin 40°
= 4.79 (cor. to 3 sig. fig.)
99
Measures, Shape and Space
Find the values of B in ABC for the following conditions. (7–10)
7.
A = 55° , a = 5.5 cm , b = 4 cm
Solution
By the sine formula,
By the sine formula,
4 sin 55°⎞
sin B = ⎛
⎝ 5.5 ⎠
B = ( 36.6° ) (cor. to 3 sig. fig.)
or
sin B
sin A
=
b
a
sin B =
6 sin 140°
12
B = 18.7° (cor. to 3 sig. fig.)
or B = 161° (rejected)
B = ( 143° ) (rejected)
A = 55° , a = 2 cm , b = 2.2 cm
10. A = 75° , a = 10 cm , b = 10 cm
Solution
Solution
By the sine formula,
By the sine formula,
sin B
sin A
=
b
a
sin B =
2.2 sin 55°
2
B = 64.3° (cor. to 3 sig. fig.)
or B = 116° (cor. to 3 sig. fig.)
100
A = 140° , a = 12 cm , b = 6 cm
Solution
sin B
sin A
=
( b )
( a )
8.
9.
sin B
sin A
=
b
a
sin B =
10 sin 75°
10
∴ B = 75° or 105° (rejected)
9
More about Trigonometry (II)
Solve ABC for the following conditions. (11–14)
11.
A = 58° , C = 80° , b = 11 cm
Solution
B = 180° − ( 58° ) − ( 80° )
= ( 42° )
By the sine formula,
sin ( 42° )
sin ( 58° )
=
( 11 )
a
a =
(
11 sin 58°
sin 42°
) cm
= ( 13.9 ) cm (cor. to 3 sig. fig.)
sin ( 42° )
sin ( 80° )
=
( 11 )
c
c = ( 11 sin 80° ) cm
sin 42°
= (
16.2
) cm (cor. to 3 sig. fig.)
12. A = 118° , B = 34° , b = 30 cm
Solution
C = 180° − 118° − 34°
= 28°
By the sine formula,
a
b
=
sin A
sin B
a =
30 sin 118°
cm
sin 34°
= 47.4 cm (cor. to 3 sig. fig.)
b
c
=
sin B
sin C
c =
30 sin 28°
cm
sin 34°
= 25.2 cm (cor. to 3 sig. fig.)
101
Measures, Shape and Space
13. B = 67° , a = 21 cm , b = 12 cm
Solution
By the sine formula,
sin A
sin B
=
a
b
sin A =
21 sin 67°
12
= 1.611
> 1
Q sin A ≤1 for all values of A
∴ The triangle does not exist.
14. A = 44° , a = 7.8 cm , c = 10.4 cm
Solution
By the sine formula,
sin C
sin A
=
c
a
sin C =
10.4 sin 44°
7.8
C = 67.9° (cor. to 3 sig. fig.)
or
112° (cor. to 3 sig. fig.)
B = 180° − A − C
= 180° − 44° − 67.852°
or
= 68.1° (cor. to 3 sig. fig.)
180° − 44° − 112.148°
or
23.9° (cor. to 3 sig. fig.)
By the sine formula,
b
a
=
sin B
sin A
b =
⎛ 7.8
⎞
× sin 68.1° cm
⎝ sin 44°
⎠
or
= 10.4 cm (cor. to 3 sig. fig) or
⎛ 7.8
⎞
× sin 23.9° cm
⎝ sin 44°
⎠
4.54 cm
∴ The solutions of ABC are:
B = 68.1°, C = 67.9°, b = 10.4 cm
or B = 23.9°, C = 112°, b = 4.54 cm
102
(cor. to 3 sig. fig.)
9
NON-FOUNDATION
More about Trigonometry (II)
Name :
9C
9.3
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Date :
Mark :
The Cosine Formula
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Key Concepts and Formulae
A
The cosine formula
a 2 = b 2 + c 2 − 2bc cos A
2
2
b + c − a
or cos A =
2 bc
c
B
b 2 = a 2 + c 2 − 2ac cos B
or cos B =
b
2
a
C
a 2 + c2 − b 2
2 ac
c 2 = a 2 + b 2 − 2ab cos C
or cos C =
a 2 + b 2 − c2
2 ab
(Unless otherwise specified, numerical answers should either be exact or correct to 3 significant
figures.)
Find the unknowns in the following triangles. (1–6)
1.
C
11 cm
x cm
A
40o
15 cm
B
Solution
By the cosine formula,
x = ( 11 ) 2 + ( 15 ) 2 − 2( 11 ) ( 15 ) cos ( 40° )
= ( 9.65 ) (cor. to 3 sig. fig.)
103
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Measures, Shape and Space
2.
4.
C
C
θ
5 cm
5 cm
7.6 cm
7 cm
A
θ
8 cm
A
Solution
Solution
By the cosine formula,
By the cosine formula,
cos θ =
2
cos θ =
=
2
( 7 ) +( 8 ) −( 5 )
2( 7 ) ( 8 )
2
5 2 + 7.6 2 − 112
2(5)(7.6)
θ = 120° (cor. to 3 sig. fig.)
( 11 )
( 14 )
θ = ( 38.2° )
(cor. to 3 sig. fig.)
D
5.
3.
B
11 cm
B
y cm
C
A
20 cm
x cm
11 cm
4 cm
40o
120o
A
73o
B
x cm
B
6 cm
C
Solution
Solution
By the cosine formula,
By the cosine formula,
202 = x 2 + 112 − 2(11) x cos 120°
x =
−11 ±
112 − 4(1)(−279)
2
x = 12.1 (cor. to 3 sig. fig.)
x = 23.1 (rejected)
x 2 = 62 + 72 − 2(6)(7) cos 73°
x =
x 2 + 11x − 279 = 0
or
7 cm
62 + 72 − 2(6)(7) cos 73°
= 7.77 (cor. to 3 sig. fig.)
By the cosine formula,
y 2 = 42 + x 2 − 2(4)(x ) cos 40°
y =
42 + 7.7742 − 2(4)(7.774) cos 40°
= 5.37 (cor. to 3 sig. fig.)
104
9
More about Trigonometry (II)
C
6.
θ
14.4 cm
y cm
x cm
B
40o
A
8 cm
9 cm
D
Solution
Consider ABC.
By the cosine formula,
x 2 = 82 + 14.42 − 2(8)(14.4) cos 40°
x =
82 + 14.42 − 2(8)(14.4) cos 40°
= 9.74 (cor. to 3 sig. fig.)
Consider BCD.
By the cosine formula,
y 2 = (8 + 9)2 + 14.42 − 2(8 + 9)(14.4) cos 40°
y =
172 + 14.42 − 2(17)(14.4) cos 40°
= 11.0 (cor. to 3 sig. fig.)
Consider ACD.
cos θ =
=
x
2
2
+ y − 9
2 xy
2
9.740 2 + 11.014 2 − 81
2(9.740)(11.014)
θ = 50.9° (cor. to 3 sig. fig.)
105
Measures, Shape and Space
Solve ABC for the following conditions. (7–8)
7.
A = 60° , b = 18 cm , c = 12 cm
Solution
By the cosine formula,
a =
=
( 18 ) 2 + ( 12 ) 2 − 2( 18 )( 12 ) cos ( 60° ) cm
( 252 ) cm
= ( 15.9 ) cm (cor. to 3 sig. fig.)
a 2 + c 2 − b2
2ac
( 252 )2 + ( 12 )2 − ( 18 )2
=
2( 252 )( 12 )
cos B =
B = ( 79.1° ) (cor. to 3 sig. fig.)
C = 180° − ( 60° ) − ( 79.1° )
= ( 40.9° ) (cor. to 3 sig. fig.)
8.
a = 8 cm , b = 10 cm , c = 13 cm
Solution
cos A =
b2 + c2 − a2
2bc
2
=
2
10 + 13 − 8
2(10)(13)
2
A = 38.0° (cor. to 3 sig. fig.)
cos B =
=
a2 + c2 − b2
2ac
8 2 + 13 2 − 10 2
2(8)(13)
B = 50.3° (cor. to 3 sig. fig.)
C = 180° − 37.958° − 50.251°
= 91.8° (cor. to 3 sig. fig.)
106
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9.
In the figure, M and N are two points on AC and AB
respectively. AM = MC = CB = 4 cm , BN = 2 cm
and NA = 7 cm .
43
(a) Show that cos A =
.
48
More about Trigonometry (II)
C
4 cm
4 cm
M
4 cm
B
2 cm N
7 cm
A
(b) Hence, or otherwise, find MN.
Solution
(a) By the cosine formula,
cos A =
AC 2 + AB 2 − BC 2
2(AC )(AB )
=
82 + 92 − 42
2(8)(9)
=
129
144
=
43
48
(b) Consider AMN.
By the cosine formula,
MN 2 = AM 2 + AN 2 − 2(AM )(AN ) cos A
MN =
⎛ 43 ⎞
42 + 72 − 2(4)(7)
cm
⎝ 48 ⎠
= 3.85 cm (cor. to 3 sig. fig.)
107
Measures, Shape and Space
10. In the figure, AB is a diameter of the circle. By using the cosine
formula, prove that d 2
Solution
Consider BCD.
cos ∠BCD =
b 2 + c 2 − BD 2
2bc
∠DAB + ∠BCD = 180°
K K (1)
(opp. ∠s, cyclic quad.)
∴ ∠DAB = 180° − ∠BCD
Consider ABD.
cos ∠DAB =
d 2 + a 2 − BD 2
2ad
cos (180° − ∠BCD ) =
d 2 + a 2 − BD 2
2ad
cos ∠BCD =
BD 2 − d 2 − a 2
2ad
K K (2)
From (1) and (2),
BD 2 − d 2 − a 2
b 2 + c 2 − BD 2
=
2ad
2bc
∴
∠ADB = 90°
K K (3)
(∠ in semi-circle)
a 2 + BD 2 = d 2
BD 2 = d 2 − a 2 K K (4)
By substituting (4) into (3), we have
d 2 − a2 − d 2 − a2
b2 + c2 − d 2 + a2
=
2ad
2bc
−2a 2
b2 + c2 − d 2 + a2
=
2ad
2bc
−2abc
= b2 + c 2 − d 2 + a2
d
d 2 = a2 + b2 + c 2 +
108
2abc
d
b
C
c
a
A
Join BD.
Q
D
2 abc
= a 2 + b 2 + c2 +
.
d
d
B
9
NON-FOUNDATION
More about Trigonometry (II)
Name :
9D
9.4
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Date :
Mark :
Heron’s Formuala
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Key Concepts and Formulae
A
Heron’s formula
Area of ABC =
s ( s − a )( s − b)( s − c) ,
c
b
a + b+ c
.
where s =
2
B
C
a
(Unless otherwise specified, numerical answers should either be exact or correct to 3 significant
figures.)
In the following figures, find the area of the triangles. (1–2)
A
1.
3.5 cm
B
2.
2.8 cm
3 cm
( 3.5 ) + ( 2.8 ) + ( 3 )
cm
( 2 )
= ( 4.65 ) cm
s − a = [( 4.65 ) − ( 3 )] cm = ( 1.65 ) cm
s − b = [( 4.65 ) − ( 2.8 )] cm = ( 1.85 ) cm
s − c = [( 4.65 ) − ( 3.5 )] cm = ( 1.15 ) cm
∴
Area of ABC
= s(s − a)(s − b)(s − c )
=
9 cm
6 cm
C
Solution
s =
C
4.65 × 1.65 × 1.85 × 1.15 cm2
= 4.04 cm2 (cor. to 3 sig. fig.)
A
5 cm
B
Solution
s =
6 + 9 + 5
cm
2
= 10 cm
s − a = (10 − 9) cm = 1 cm
s − b = (10 − 6) cm = 4 cm
s − c = (10 − 5) cm = 5 cm
∴ Area of ABC
=
s(s − a)(s − b)(s − c )
=
10 × 1 × 4 × 5 cm2
= 14.1 cm2 (cor. to 3 sig. fig.)
109
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Measures, Shape and Space
In the following figures, the altitude of triangle is h cm. Find the value of h. (3–4)
3.
D
4 cm
4.
C
24 cm
2.5 cm
h cm
h cm
A
4.5 cm
14 cm
B
C
A
12 cm
D
B
Solution
Solution
s =
s =
4 + 2.5 + 4.5
cm
2
= 25 cm
Area of ABC
= 5.5 cm
Area of ACD
=
5.5 × (5.5 − 4.5) × (5.5 − 4) × (5.5 − 2.5) cm2
=
24.75 cm2
Q
1
× AC × BD = area of ACD
2
∴
1
× 4.5 × h =
2
24.75
h = 2.21 (cor. to 3 sig. fig.)
110
12 + 14 + 24
cm
2
=
25 × (25 − 12) × (25 − 14) × (25 − 24) cm2
=
3575 cm2
Q
1
× AB × CD = area of ABC
2
∴
1
× 12 × h =
2
3575
h = 9.97 (cor. to 3 sig. fig.)
9
5.
The figure shows a cyclic quadrilateral ABCD. If AB = 13 cm ,
BC = 5 cm , CD = 7.5 cm , DA = 6.5 cm and AB is a
diameter of the circle, find the area of quadrilateral ABCD.
D
7.5 cm
6.5 cm
C
5 cm
A
Solution
More about Trigonometry (II)
13 cm
B
Join AC.
∠ACB = 90° (∠ in semi-circle)
By Pythagoras’ theorem,
AC =
=
AB 2 − CB 2 cm
132 − 52 cm
= 12 cm
1
∴ Area of ABC = ⎛ × 5 × 12⎞ cm2
⎝2
⎠
= 30 cm2
Consider ACD.
By Heron’s formula,
s =
6.5 + 7.5 + 12
cm
2
= 13 cm
∴ Area of ACD =
13 × (13 − 12) × (13 − 6.5) × (13 − 7.5) cm2
= 21.6 cm2 (cor. to 3 sig. fig.)
∴ The area of quadrilateral ABCD = (30 + 21.6) cm2
= 51.6 cm2
111
Measures, Shape and Space
6.
In the figure, OAC is a sector of radius 16 cm with centre O.
If AB = 10 cm and BC = 20 cm , find the area of the
shaded region.
C
20 cm
16 cm
Solution
10 cm
Join OB.
O
Consider OAB.
By the cosine formula,
cos ∠AOB =
16 2 + 16 2 − 10 2
2(16)(16)
∠AOB = 36.4199°
∴
By Heron’s formula,
s =
16 + 16 + 10
cm
2
= 21 cm
Area of OAB =
21 × (21 − 16) × (21 − 16) × (21 − 10) cm2
5775 cm2
=
Consider OBC.
By the cosine formula,
cos ∠BOC =
16 2 + 16 2 − 20 2
2(16)(16)
∠BOC = 77.3644°
∴
By Heron’s formula,
s =
16 + 16 + 20
cm
2
= 26 cm
26 × (26 − 16) × (26 − 16) × (26 − 20) cm2
Area of OBC =
15 600 cm2
=
Consider the sector OAC,
area of the sector =
77.3644° + 36.4199°
× π × 162
360°
= 254.197 cm2
∴ The area of the shaded region = (254.197 −
5775 −
15 600 ) cm2
= 53.3 cm2 (cor. to 3 sig. fig.)
112
B
A
9
NON-FOUNDATION
More about Trigonometry (II)
Name :
9E
9.5
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Mark :
Trigonometric Problems in Two Dimensions
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(Unless otherwise specified, numerical answers should either be exact or correct to 3 significant
figures.)
1.
The figure shows three buildings A, B and C. The true bearings
of B from C is 100° and C from A is 050°. It is known that
CB = 7.5 km and CA = 5 km. How far is B from A?
N
Solution
Mark the angles α and β as shown in the figure.
5 km
50o
N
o
C 100
α
β
7.5 km
B
A
α = 50°
β = 180° − 100°
(adj. ∠s on st. line)
= 80°
∴
∠ACB = 50° + 80°
= 130°
Consider ABC.
By the cosine formula,
AB =
=
AC 2 + BC 2 − 2(AC )(BC ) cos ∠ACB
52 + 7.52 − 2(5)(7.5) cos 130° km
= 11.4 km (cor. to 3 sig. fig.)
113
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Measures, Shape and Space
2.
The angles of elevation of the top D of a building from the
bottom B and the top A of another building are 40° and 30°
respectively. If the height of AB is 70 m, find the height of
building CD.
D
A
Solution
70 m
Let h m be the height of building CD.
30o
40o
C
B
h
h − 70
=
tan 40°
tan 30°
h tan 30° = h tan 40° − 70 tan 40°
h =
70 tan 40°
tan 40° − tan 30°
= 224 (cor. to 3 sig. fig.)
∴ The height of building CD is 224 m.
3.
N
A man at a lighthouse L, observed two ships, P and Q. The true bearing
of P from L is 040°. The distance between P and Q was 5 km, and the
true bearing of Q from L was 150°. It is known that ∠LQP = 40° . Find
(a) the true bearing of L from P,
(b)
P
α
the true bearing of L from Q,
(c) the distance between P and L.
N
40o
Solution
5 km
With the notations in the figure,
(a)
α = 40° . The true bearing of L from P = 180° + 40° = 220°
(b)
β = 150° − 90° = 60° . The true bearing of L from Q = 270° + 60° = 330°
(c)
∠PLQ = 150° − 40°
= 110°
By the sine formula,
PL
PQ
=
sin ∠PQL
sin ∠PLQ
PL =
5
⎛
⎞
× sin 40° km
⎝ sin 110°
⎠
= 3.42 km (cor. to 3 sig. fig.)
114
L
150o
N
40o
β
Q
9
4.
From the top A of a flagstaff AB, the
angles of depression of the top C and the
bottom D of another flagstaff CD are 30°
and 60° respectively. If the height of
flagstaff AB is 80 m and the height of
flagstaff CD is h m, find the value of h.
A
30o
5.
More about Trigonometry (II)
In the figure, AB is a tree and CD is a
building. The angle of elevation of A
from D is 15°, while the angle of
depression of A from C is 35°. If the
height of the building is 40 m, find the
height of the tree.
F
F
C
35o
60o
C
E
40 m
80 m
hm
E
A
15o
B
D
With the notations in the figure,
∠ADB = 60° (alt ∠s, AF // BD)
AE =
Consider AEC.
∠ACE = 30° (alt ∠s, AF // EC)
80 − h
m
tan 30°
Q BD = EC
80
80 − h
∴
=
tan 60°
tan 30°
3
Consider AEC.
∠CAE = 35° (alt ∠s, CF // AE)
80 m
BD =
tan 60°
=
With the notations in the figure,
let h m be the height of the tree.
consider ABD.
80
D
Solution
Solution
EC =
B
Consider ABD.
BD =
40 − h
h
=
tan 35°
tan 15°
40 tan 15° − h tan 15° = h tan 35°
h =
80
3
= 53.3 (cor. to 3 sig. fig.)
h
m
tan 15°
Q AE = BD
3 (80 − h)
h = 80 −
40 − h
m
tan 35°
40 tan 15°
tan 15° + tan 35°
= 11.1 (cor. to 3 sig. fig.)
∴ The height of the tree is 11.1 m.
115
Measures, Shape and Space
6.
(b) Let h km be the required shortest distance.
The figure shows three towns A, B and
C. The true bearing of B from C is 135°,
AB = 52 km , BC = 38 km and
AC = 25 km .
1
1
× AB × h =
× AB × BC × sin ∠ABC
2
2
∴ h = BC sin ∠ABC
= (38 sin 26.944°) km
N
o
135
C
= 17.2 km (cor. to 3 sig. fig)
∴ The shortest distance of the man
25 km
h km
A
from town C is 17.2 km.
38 km
N
52 km
D
B
(a) Find the compass bearing of A from B.
(b) If a man travels from B to A, what
is the shortest distance of the man
from town C.
7.
In the figure, CD is a lamppost standing
vertically on a hillside AC sloping at 20°
to the horizontal. E is a point on AC such
that E is 3 m from C. The angle of
elevation of D from E is 65°. Find the
height of the lamppost CD.
D
Solution
(a) With the notations in the figure,
by the cosine formula,
65o
AB 2 + BC 2 − AC 2
cos ∠ABC =
2( AB )(BC )
=
∴
52 2 + 38 2 − 25 2
2(52)(38)
∠ABC = 26.944°
E
3m
C
20o
A
B
Solution
∠DBA = (180° − 135°) + 26.944°
= 71.9° (cor. to 3 sig. fig.)
∴ The compass bearing of A from
B is N71.9°W .
Consider CDE.
∠DEC = 65° − 20° = 45°
∠EDC = 180° − 90° − 65°
= 25°
By the sine formula,
CD
EC
=
sin ∠DEC
sin ∠EDC
3m
CD
=
sin 45°
sin 25°
CD = 5.02 m (cor. to 3 sig. fig.)
∴ The height of the lamppost is 5.02 m.
116
9
8.
N
40
km
E
D
A
/h
/h
N
40o
60o
km
(b) the compass bearing of C from D
P
B
25
Two cars A and B leave P at the same time. Car A travels at 25 km/h
on a course of S40°E and car B travels at 40 km/h on a course of
S60°W. Find
(a) the distance between the two cars,
More about Trigonometry (II)
C
after two hours.
Solution
(a) After two hours, the distance travelled by car A = (2 × 25) km = 50 km; the distance
travelled by car B = (2 × 40) km = 80 km
By the cosine formula,
DC =
PC 2 + PD 2 − 2(PC )(PD ) cos ∠DPC
=
502 + 802 − 2(50)(80) cos 100° km
= 101 km (cor. to 3 sig. fig.)
∴ The distance between the two cars is 101 km after two hours.
(b) By the sine formula,
DC
PC
=
sin ∠DPC
sin ∠PDC
101.44
sin 100°
=
sin ∠PDC =
50
sin ∠PDC
50
× sin 100°
101.44
∠PDC = 29.0° (cor. to 3 sig. fig.)
With the notation in the figure,
∠EDC = 60° + 29.0°
= 89.0°
∴ The compass bearing of C from D is N89.0°E after two hours.
117
9
NON-FOUNDATION
More about Trigonometry (II)
Name :
9F
Date :
Mark :
9.6B Problems in Three Dimensions
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Key Concepts and Formulae
Line of greatest slope
X
F
If PX ⊥ AB , PX is the line of greatest slope of the
inclined plane ABEF.
A
E
D
P
C
B
(Unless otherwise specified, numerical answers should either be exact or correct to 3 significant
figures.)
1.
H
The figure shows a cube of side 20 cm. Find
(a) the length of BH,
G
E
F
(b) the angle between the line BH and the plane ABCD.
D
Solution
(a) Join BD and consider ABD.
( BD )2 = ( AB )2 + ( AD )2
A
B
(Pyth. theorem)
Consider HBD.
BH 2 = ( BD ) 2 + ( HD ) 2
2
2
(Pyth. theorem)
∴ BH = ( AB ) + ( AD ) + ( HD ) 2
2
∴ BH = 202 + 202 + 202 cm
= 34.6 cm (cor. to 3 sig. fig.)
(b) Let θ be the angle between BH and the plane ABCD.
sin θ =
( HD
( BH
)
(
)
20
=
)
( 34.641 )
∴ θ = ( 35.3° ) (cor. to 3 sig. fig.)
∴
118
C
θ
The angle between the line BH and the plane ABCD is (
35.3°
).
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9
2.
H
In the figure, ABCDEFGH is a rectangular block with
AB = 12 cm , BC = 9 cm and CG = 5 cm . Find
(a) the length of AG,
More about Trigonometry (II)
G
5 cm
E
F
D
(b) the angle between the line AG and the plane BCGF.
C
9 cm
Solution
A
(a) Join AC and AG.
12 cm
B
Consider ABC.
AC 2 = AB 2 + BC 2
(Pyth. theorem)
Consider ACG.
AG 2 = AC 2 + CG 2
(Pyth. theorem)
∴ AG 2 = AB 2 + BC 2 + CG 2
AG =
=
122 + 92 + 52 cm
250 cm
= 15.8 cm (cor. to 3 sig. fig.)
(b) The angle between AG and the plane BCGF is ∠AGB.
BG =
BC 2 + CG 2
=
92 + 52 cm
=
106 cm
tan ∠AGB =
=
(Pyth. theorem)
AB
BG
12
106
∠AGB = 49.4° (cor. to 3 sig. fig.)
∴ The angle between AG and the plane BCGF is 49.4°.
119
Measures, Shape and Space
3.
In the figure, ABCDEFGH is a rectangular block with AB = 8 cm ,
BC = 6 cm and CG = 5 cm . Find the angles between
E
(a) AG and plane ADHE,
H
G
F
5 cm
D
(b) HB and plane CGHD.
A
C
6 cm
8 cm
B
Solution
(a) The angle between AG and the plane ADHE is ∠GAH.
AH =
=
=
AD 2 + DH 2
2
(Pyth. theorem)
H
6 + 5 cm
5 cm
D
61 cm
tan ∠GAH =
F
E
2
G
A
C
6 cm
8 cm
B
8
61
∴ ∠GAH = 45.7° (cor. to 3 sig. fig.)
∴ The angle between AG and the plane ADHE is 45.7°.
(b) The angle between HB and the plane CGHD is ∠BHC.
HC =
2
DC + HD
2
2
(Pyth. theorem)
=
8 + 5 cm
=
89 cm
=
89
∴ ∠BHC = 32.5° (cor. to 3 sig. fig.)
∴ The angle between HB and the plane CGHD is 32.5°.
5 cm
D
A
6
G
F
E
2
BC
tan ∠BHC =
HC
120
H
C
6 cm
8 cm
B
9
4.
More about Trigonometry (II)
T
The figure shows a right pyramid of height 8 cm. Its base is a
square of side 10 cm. Find
(a) the length of TQ,
S
(b) the angle between the plane TQR and the plane PQRS.
Solution
OQ =
1
SQ
2
P
10 cm
Q
(Property of square)
200
cm
2
=
50 cm
=
(Pyth. theorem)
200 cm
=
TQ =
M
O
(a) SQ = 102 + 102 cm
=
R
8 cm
82 + ( 50 )2 cm
114 cm
= 10.7 cm (cor. to 3 sig. fig.)
(b) QR is the line of intersection of the planes TQR and PQRS.
Let M be the mid-point of QR.
Q TQR is an isosceles triangle.
∴ TM ⊥ QR
Q OQR is also an isosceles triangle.
∴ OM ⊥ QR
The angle between the planes TQR and PQRS is ∠TMO.
Consider TOM.
tan ∠TMO =
TO
8
=
MO
5
∠TMO = 58.0° (cor. to 3 sig. fig.)
∴ The angle between the planes TQR and PQRS is 58.0°.
121
Measures, Shape and Space
5.
In the figure, ABCDEF is a right triangular prism. If CF = 15 cm ,
AB = BC = 5 cm and AC = 6 cm , find the angle between the
planes ACE and ABC.
D
F
E
Solution
15 cm
AC is the line of intersection of the planes ACE and ABC.
By symmetry, AE = EC .
AE =
=
∴
2
5 + 15 cm
5 cm
250 cm
B
AE = EC =
250 cm
Let M be the mid-point of AC.
Q ACE is an isosceles triangle
∴ EM ⊥ AC
Q ABC is an isosceles triangle
∴ MB ⊥ AC
∴ The angle between the plane ACE and the plane ABC is ∠EMB.
EM =
AE 2 − AM 2
=
250 − 32 cm
=
241 cm
BM =
AB 2 − AM 2
=
52 − 32 cm
= 4 cm
Consider EBM.
cos ∠EMB =
EM
2
2
+ MB − EB
2(EM )(MB)
2
=
241 + 4 − 15
2
2
2( 241)(4)
∠EMB = 75.1° (cor. to 3 sig. fig.)
∴ The angle between the planes ACE and ABC is 75.1°.
122
6 cm
C
M
5 cm
A
2
9
6.
The figure shows a hillside ABFE sloping at 30° to the
horizontal ground ABCD. AF is a straight path of 100 m
long, making an angle 40° with AB. Find
F
E
C
D
(a) the height of E above the ground,
0m
10
(b) the inclination of the path AF to the ground.
Solution
More about Trigonometry (II)
30o
θ
A
40o
B
(a) BF = AF sin 40°
= 100 sin 40° m
FC = BF sin 30°
= (100 sin 40° × sin 30°) m
= 32.1 m (cor. to 3 sig. fig.)
Q ED = FC
∴ The height of E above the ground is 32.1 m.
(b) Let θ be the angle of inclination of the path AF to the ground.
sin θ =
=
FC
AF
32.139
100
∴ θ = 18.7° (cor. to 3 sig. fig.)
∴ The inclination of the path AF to the ground is 18.7°.
123
Measures, Shape and Space
7.
A door of dimensions 2 m by 1 m is opened through an angle of 40°
from position AEFD to position ABCD as shown. Find
D
F
40o
C
(a) the length of AC,
2m
(b) the distance between C and F,
θ
(c) the angle between AC and AF.
A
Solution
1m
AB 2 + BC 2
(a) AC =
(Pyth. theorem)
12 + 22 m
=
= 2.24 m (cor. to 3 sig. fig.)
(b) By the cosine formula,
DC 2 + DF 2 − 2(DC )(DF ) cos ∠FDC
CF =
12 + 12 − 2(1)(1) cos 40° m
=
= 0.684 m
(cor. to 3 sig. fig.)
(c) Let θ be the angle between AC and AF.
Consider ACF.
cos θ =
AC
2
2
+ AF − CF
2( AC )(AF )
2
=
∴
2
2
( 5 ) + ( 5 ) − 0.6840
2
2( 5 )( 5 )
θ = 17.6° (cor. to 3 sig. fig.)
∴ The angle between AC and AF is 17.6°.
124
E
B
9
8.
The figure shows a cuboid of
dimensions 3 cm × 5 cm × 7 cm . The
diagonal AG makes an angle θ with the
face BCGF. Find the angle θ.
H
G
9.
More about Trigonometry (II)
The figure shows a right pyramid
VABCD with a square base of side 10 cm.
If VA = VB = VC = VD = 15 cm ,
VM ⊥ AB and VN ⊥ CD, find the angle
between the lines VM and VN.
V
θ
E
7 cm
15 cm
F
D
C
D
C
N
5 cm
A
10 cm
B
3 cm
A
AC 2 = AB 2 + BC 2
(Pyth. theorem)
AG 2 = AC 2 + CG 2
∴ AG
2
2
(Pyth. theorem)
2
= AB + BC + CG
∴ AG =
=
BG =
2
Solution
The angle between the lines VM and VN
is ∠MVN.
32 + 52 + 72 cm
Clearly, AM = MB
83 cm
∴ AM =
BC 2 + CG 2
2
(Pyth. theorem)
2
= 5 cm
5 + 7 cm
Consider VAM.
=
74 cm
VM =
By the cosine formula,
cos θ =
AG
+ BG
=
2
=
− AB
2
2
( 83 ) + ( 74 ) − 3
152 − 52 cm
(Pyth. theorem)
200 cm
By symmetry, VM = VN =
2( AG )(BG )
2
∴
10 cm
2
=
2
B
M
Solution
200 cm .
MN = BC
2
2( 83 )( 74 )
θ = 19.2° (cor. to 3 sig. fig.)
= 10 cm
Consider VMN.
By the cosine formula,
cos ∠MVN =
VM 2 + VN 2 − MN 2
2(VM )(VN )
2
=
2
( 200 ) + ( 200 ) − 10
2
2( 200 )( 200 )
= 0.75
∴
∠MVN = 41.4° (cor. to 3 sig. fig.)
125
Measures, Shape and Space
H
10. The figure shows a cube of side 10 cm. The diagonals EC and
BH intersect at K. Find the angle θ.
G
E
F
Solution
Consider rectangle EBCH.
HK = KB
(property of rectangle)
EK = KC
(property of rectangle)
AC 2 = AB 2 + BC 2
(Pyth. theorem)
EC 2 = AC 2 + AE 2
(Pyth. theorem)
∴ EC 2 = AB 2 + BC 2 + AE 2
EC =
=
102 + 102 + 102 cm
300 cm
By symmetry,
EC = HB
1
EC
2
∴ KB = KC =
=
300
cm
2
=
75 cm
Consider KBC.
By the cosine formula,
cos θ =
KB 2 + KC 2 − BC 2
2(KB )(KC )
2
=
=
∴
126
( 75)
+ ( 75 )2 − 10 2
2( 75 )( 75 )
1
3
θ = 70.5° (cor. to 3 sig. fig.)
K
10 cm
θ
D
C
10 cm
A
10 cm
B
9
NON-FOUNDATION
More about Trigonometry (II)
Name :
9G
Date :
Mark :
9.6C Further Applications of Trigonometry in Three Dimensions
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(Unless otherwise specified, numerical answers should either be exact or correct to 3 significant
figures.)
1.
T
From a point B due south of a flagstaff TA, a man observes that the
angle of elevation of T is 20°. He then walks 200 m in the direction
N65°W to a point C, which is in the direction S55°W from A. Find
N
(a) the height of the flagstaff TA,
(b) the angle of elevation of T from C.
Solution
(a)
A
C
55o
200 m
∠ACB = 180° − 65° − 55° (∠ sum of )
= 60°
20o
65o
B
By the sine formula,
AB
CB
CA
=
=
sin ∠ACB
sin ∠CAB
sin ∠CBA
200
AB
CA
=
=
sin 60°
sin 55°
sin 65°
∴ AB =
200 sin 60°
m
sin 55°
CA =
200 sin 65°
m
sin 55°
Consider TAB.
TA
= tan 20°
AB
∴ TA =
⎛ 200 sin 60°
⎞
× tan 20° m
⎝ sin 55°
⎠
= 77.0 m (cor. to 3 sig. fig.)
127
○
○
Measures, Shape and Space
(b) Consider TAC.
tan ∠TCA =
=
TA
CA
sin 55°
200 sin 60° tan 20°
×
sin 55°
200 sin 65°
∴ ∠TCA = 19.2° (cor. to 3 sig. fig.)
∴ The angle of elevation of T from C is 19.2°.
2.
In the figure, ABC is located on the horizontal ground.
T is vertically above C. If AB = 150 m , ∠CAB = 30° ,
∠CBA = 35° and ∠TBC = 40° , find
T
(a) the height of TC,
C
(b) the angle of depression of A from T.
A
Solution
(a) ∠ACB = 180° − 30° − 35°
(∠ sum of )
= 115°
By the sine formula,
BC
sin 30°
=
∴ BC =
AC =
AC
sin 35°
=
150
sin 115°
150 sin 30°
m
sin 115°
150 sin 35°
sin 115°
m
Consider TBC.
TC
= tan 40°
BC
TC =
⎛ 150 sin 30°
⎞
× tan 40° m
⎝ sin 115°
⎠
= 69.4 m (cor. to 3 sig. fig.)
(b) Q The angle of depression of A from T is equal to ∠TAC.
Consider TAC.
tan ∠ TAC =
=
∴
128
TC
AC
150 sin 30° tan 40°
sin 115°
×
sin 115°
150 sin 35°
∠TAC = 36.2° (cor. to 3 sig. fig.)
∴ The angle of depression of A from T is 36.2°.
40o
35o
30o
150 m
B
9
3.
Two ships X and Y are due east and due south of a lighthouse
TM of height 50 m respectively. The angles of elevation of T
from X and Y are 30° and 45° respectively. Find
T
N
50 m
N
(a) the compass bearing of Y from X,
(b) the distance between X and Y.
More about Trigonometry (II)
W
30o
E
45o
Solution
P
Y
(a) With the notations in the figure, consider TMX.
MX =
X
M
50
tan 30°
S
m = 50 3 m
Consider TMY.
MY =
50
m = 50 m
tan 45°
Consider XYM.
tan ∠YXM =
=
=
MY
MX
50
50 3
1
3
∠YXM = 30°
∠PXY = 90° − 30° = 60°
∴ The compass bearing of Y from X is S60°W.
(b) XY = MX 2 + MY 2
=
(Pyth. theorem)
(50 3 )2 + 502 m
= 100 m
∴ This distance between X and Y is 100 m.
129
Measures, Shape and Space
4.
A, B and C are three points on a straight line which lies on a
horizontal plane such that AB = BC = 30 m . TO is a
vertical tower of height 25 m. The foot O of the tower lies on
the same plane as A, B and C. If A and C are due south and
due east of O respectively, and the angle of elevation of T
from A is 40°, find
(a) the compass bearing of A from C,
(b) the angle of elevation of T from C,
(c) the angle of depression of B from T.
Solution
(a) With the notations in the figure,
AO =
25
m
tan 40°
= 29.7938 m
Consider OAC.
sin ∠OCA =
=
AO
AC
25
60 tan 40°
∠OCA = 29.7729°
θ = 90° − 29.7729° = 60.2° (cor. to 3 sig. fig.)
∴ The compass bearing of A from C is S60.2°W.
(b) OC =
=
AC 2 − OA2
(Pyth. theorem)
602 − 29.79382 m
= 52.0800 m
tan ∠TCO =
25
52.0800
∴ ∠TCO = 25.6° (cor. to 3 sig. fig.)
∴ The angle of elevation of T from C is 25.6°.
130
T
25 m
N
C
N
E
O
o
40
30 m
A
B
30 m
θ
9
More about Trigonometry (II)
(c) The angle of depression of B from T is equal to ∠TBO.
Consider OBC.
By the cosine formula,
OB 2 = BC 2 + OC 2 − 2(BC )(OC ) cos ∠OCA
OB =
302 + 52.08002 − 2(30)(52.0800) cos 29.7729° m
= 29.9999 m
Consider TOB.
tan ∠TBO =
=
TO
OB
25
29.9999
∴ ∠TBO = 39.8° (cor. to 3 sig. fig.)
∴ The angle of depression of B from T is 39.8°.
5.
From a point A due west of a tower TC, which is of height
400 m, the angle of elevation of T is θ. From another point B
due south of the tower, the angle of elevation of T is 30°.
If AB = 800 m ,
T
400 m
N
A
θ
(a) find BC,
C
E
800 m
(b) express AC in terms of θ,
30o
(c) by considering ABC, find θ.
B
Solution
(a) tan ∠TBC =
∴
BC =
TC
BC
400 m
tan 30°
= 400 3 m
= 693 m (cor. to 3 sig. fig.)
(b)
tan θ =
∴
AC =
=
TC
AC
TC
tan θ
400
m
tan θ
131
Measures, Shape and Space
(c) Q ∠ACB = 90°
AB 2 = AC 2 + BC 2
∴
8002 =
⎛ 400 ⎞
⎝ tan θ ⎠
⎛ 400 ⎞
⎝ tan θ ⎠
2
+ (400 3 )2
2
= 8002 − (400 3 )2
tan2θ =
400 2
800 2 − ( 400 3 )2
= 1
tan θ = 1
θ = 45°
6.
In the figure, VZ is a tower of height x m due north of a point X.
Y is a point 250 m in the direction of S55°E from X. The angles
of elevation of V from X and Y are 50° and 40° respectively.
Find the height of the tower.
V
N
xm
Z
Solution
50o
x
x
m, ZY =
m
tan 50°
tan 40°
XZ =
= 125°
ZY 2 = XZ 2 + XY 2 − 2(XZ )(XY ) cos 125°
2
2
tan 40
=
x
2
⎛ x ⎞
+ 250° − 2
(250) cos 125°
⎝ tan 50° ⎠
tan 50
2
1 ⎞ 2
⎛ 1
⎛ 500
⎞
x +
−
cos 125° x − 2502 = 0
2
⎝ tan 2 40
⎠
⎝
⎠
tan 50°
tan 50
0.7162x 2 − 240.64x − 2502 = 0
By the quadratic formula,
x = 508 (cor. to 3 sig. fig.) or x = −172 (rejected)
∴ The height of the tower is 508 m.
132
250 m
S
By the cosine formula,
E
40o
55o
(adj. ∠s on st. line)
∠ZXY = 180° − 55°
x
X
W
Y
9
NON-FOUNDATION
More about Trigonometry (II)
Name :
9H
Date :
Mark :
Multiple Choice Questions
1.
In the figure AB = 2 x , AC = x
and ∠BAC = 30° . If the area of
¡ ABC = 8, find the value of x.
3.
Find the area of the parallelogram as
shown in the figure.
D
C
2x
A
30o
B
6 cm
120o
x
A
C
A. 2
A. 22.1 cm2
B.
3
B.
27.3 cm2
C.
4
C.
28.5 cm2
C
D. 5
2.
In the figure, if AC = 4 cm ,
∠CAB = 105° and ∠ACB = 30° , find
BC.
B
10.5 cm
D
D. 54.6 cm2
4.
In the figure,
AC
=
AB
C
C
30o
75o
4 cm
A
45o
B
o
105
A
B
A.
3
.
3
B.
3
.
2
C.
6
.
2
D.
6
.
3
A. 4.4 cm
B.
4.8 cm
C.
5.1 cm
D. 5.5 cm
D
D
133
Measures, Shape and Space
5.
In the figure, x =
A.
6 2m
C
B.
6( 3 + 1) m
C.
12 3 m
D.
18(1 +
2x cm
x cm
30o
A
6.
B
10 cm
A.
5.
B.
6.
C.
10 3
.
3
D.
15 3
.
3
8.
θ
C
A
A
50o
A.
B.
7.
9.
A.
tan θ = 2 tan φ
B.
tan θ = tan 2φ
C.
tan θ = 4 tan φ
D.
tan θ − tan φ =
C
C
2
In the figure, the area of ¡ ABC =
D
C
3 cm
5 sin 70°
cm
sin 60°
70o
3 sin 70° cm
C.
5 sin 50°
cm
sin 70°
D.
2 sin 85° cm
A
A
In the figure, ∠ABD = 60° ,
∠BCD = 45° and BC = 12 m .
Find AB.
60o
A
φ
B
C
35o
D
134
In the figure, BC = 3 AB . Which of the
following MUST be true?
D
In the figure, AC = CB = 5 cm ,
∠ABC = 35° and ∠CAD = 50° .
Find AD.
B
B
2) m
45o
B 12 m
C
A.
9 sin 2 70°
cm 2
2 sin 40°
B.
6 sin 110° cm 2
C.
3 sin 40°
cm 2
sin 70°
D.
3 sin 2 40°
cm 2
2 sin 70°
40o
B
A
9
10. In the figure, the compass bearings of B
and C from A are N70°E and N30°E
respectively. The compass bearing of C
from B is N35°W. If AB = 200 m , then
BC =
More about Trigonometry (II)
12. In the figure, OP is a pole which is
perpendicular to the plane OAB. The
angles of elevation of P from A and B are
30° and 45° respectively. Find the height
of pole OP.
P
C
N
N
35o
45o
N
B
30o
30o
E
80 m
A
200 m
B
O
70o
A
A. 40 m
A. 140 m.
B.
142 m.
C.
150 m.
B.
45 m
C.
50 m
A
D. 62 m
D. 172 m.
B
11. In the figure, VC is a vertical pole
standing on the horizontal ground ABC.
If ∠VBC = 60° , ∠VAC = 45° and
∠ABC = α , find tan α.
13. In the figure, ABCDEFG is a cube. Find
cos θ .
H
G
θ
E
F
V
N
45o
C
60o
α
D
A
B
A. 1
B.
1
C.
3
B.
3
C.
2 3
D. 2
B
E
A
A.
C
D.
1
2
1
3
2
3
D
B
135
Measures, Shape and Space
14. In the figure, the true bearing of Q from
P is
N
R
55o
30 m
16. In the figure, A, B and C are located on
the horizontal ground. X is 60 m
vertically above C. If AB = 180 m ,
∠CAB = 36° and ∠CBA = 40° , find
the angle of depression of A from X.
86 m
P
X
96 m
60 m
Q
C
A. 101.2°
B.
113.5°
C.
116.8°
A
40o
B
180 m
A. 20°
C
D. 123.1°
15. In the figure, the compass bearing of C
from A is N30°E while the compass
bearing of A from B is N75°W. If
AC = 40 m and AB = 70 m , find BC.
N
36o
C
B.
21.5°
C.
25.3°
D
D. 26.7°
17. The figure shows a right triangular
prism. Find the angle between AC and
the plane BCEF, correct to the nearest
degree.
30o
A
40 m
D
N
1 cm
o
75
A
3 cm
B
B
A. 69.2 m
B.
71.1 m
C.
76.8 m
D. 80.5 m
136
E
F
70 m
4 cm
C
A. 11°
B
B.
12°
C.
13°
D. 14°
A
9
18. In the figure, AB and CD are two vertical
flagstaffs standing on the horizontal
ground. The angle of elevation of D
from B is 20° while the angle of
depression of D from A is 25°. Find BD.
A
More about Trigonometry (II)
20. A man cycles from P to Q at 10 km/h in
the direction of S50°W in half an hour.
He then cycles at the same speed to R
which is due south of P. How long does
the man take to travel from Q to R if R is
in the direction of S30°E from Q.
25o
N
D
4m
N
20o
B
P
50o
Q
C
30o
A. 2.1 m
B.
3.5 m
C.
4.6 m
19. In the figure, express BC in terms of x,
α, β and γ.
C
A. 0.52 h
B.
0.68 h
C.
0.77 h
D. 0.78 h
β
D
R
D
D. 5.1 m
C
α
x
γ
A
B
A. x sin α sin β
B.
x sin α
sin β sin γ
C.
x sin 2α
sin β
D.
x sin γ
sin γ sin β
B
137