10/12/2015
Chapter 11 – Trigonometric (Applications)
Q1
The bearing of B from A is S 30o W
The bearing of A from B is N 30o E
N
A
30o
Chapter 11 – Trigonometric (Applications)
Q2
b
3
= sin 60o =
, a :b = 2: 3
a
2
c
1
= sin 30o = , a : c = 2 : 1
a
2
a : b : c = 2 : 3 :1
N
30o
B
Option A
Option C
Chapter 11 – Trigonometric (Applications)
Q3
a
= tan θ , a = p tan θ
p
With the notation in the figure
a = r sin( 270o − α ) = − r cos α
b = q − a = q − p tan θ
b
= tan θ
CD
b
q
CD =
=
−p
tan θ tan θ
Option E
Chapter 11 – Trigonometric (Applications)
Q4
The required distance = r − r cos α = r (1 − cos α )
270o - α
θ
b
a
Option C
a
r
Chapter 11 – Trigonometric (Applications)
Q5
r
Chapter 11 – Trigonometric (Applications)
Q6
π c = 180o
o
x
 180 x 
x c = ×180 o = 

π
 π 
With the notation in the figure
PC = (a + 1) cosθ , QC = a cosθ
PQ = (a + 1) cosθ − a cosθ = coaθ
Option A
Option B
a
1
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q7
Chapter 11 – Trigonometric (Applications)
Q8
With the notation in the figure
2
2
2
2
With the notation in the figure
a
b
= cos θ , a = p cos θ , = sin φ , b = q sin φ
p
q
2
a = 1 + 1 = 2, b = a + 1 = 3 , c = b + 1 = 2
cosθ =
3
2
AB = a + b = p cosθ + q sin φ
c
b
Option D
Option D
a
a
Chapter 11 – Trigonometric (Applications)
Q9
b
Chapter 11 – Trigonometric (Applications)
Q10
With the notation in the figure
a
= sin α , a = h sin α
h
a
a
h sin α
= tan β , CD =
=
CD
tan β
tan β
With the notation in the figure
16
a=
=8
2
102 − a 2 3
cosθ =
= = 0.6
10
5
Option B
Option E
a
a
Chapter 11 – Trigonometric (Applications)
Q11
Chapter 11 – Trigonometric (Applications)
Q12
With the notation in the figure
h
θ = 20o , = tan 20o
a
h
a=
m
tan 20 o
With the notation in the figure
a
= tan β , a = r tan β
r
a
a
r tan β
= cos α , AC =
=
AC
cos α
cos α
20o
h
Option E
Option E
a
θ
a
2
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q14
Chapter 11 – Trigonometric (Applications)
Q13
q− p
= cos θ
BC
q− p
BC =
cos θ
With the notation in the figure
d
= tan 45o , a = d
a
d −h
h
= tan θ , d − h = d tan θ , d =
a
1 − tan θ
A
p
B
θ
D
C
q
q–p
Option E
h
Option E
a
Chapter 11 – Trigonometric (Applications)
Q15
BC
= cos θ
a
CD
= sin θ
BC
CD
= cos θ sin θ
a
CD = a cos θ sin θ
Option D
Chapter 11 – Trigonometric (Applications)
Q16
1
( AB )( AC ) sin 30o = 64 cm 2
2
( AB ) 2 = 256 cm 2
A
a
B
θ
θ
p−q
tan θ
2
The required area
1
= × ( p + q) × h
2
( p + q )( p − q )
=
tan Aθ
1 2 4 2
= ( p − q ) tan θ
h
4
Option D
θ
h
AB = 16 cm
B
C
D
Option C
Chapter 11 – Trigonometric (Applications)
Q18
a
AB
=
sin(180 o − 20 0 − 50 o ) sin 50 o
= tan θ , h =
D
30o
C
Chapter 11 – Trigonometric (Applications)
Q17
p −q
2
A
AB =
q
a sin 50o
a sin 50o
=
sin(180o − 70o ) sin 70o
B
Option E
θ
p
C
3
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q19
GB
= cos θ , GB = q cos θ
q
BF
= sin θ , BF = p sin θ
p
C G
D
q
CE = GB + BF
= p sin θ + q cos θ
Option E
Chapter 11 – Trigonometric (Applications)
Q20
p
X
θ
E F Y
A
h
= tan 45o , OC = h
OC
h
= tan 30o , OB = 3h
OB
OC
= tan θ
B
OB
1
o
tan θ =
= tan 30
3
Option B
3
θ
Option A
Chapter 11 – Trigonometric (Applications)
Q23
x 2 = 32
x = 32 = 5.66
Option D
B
C
D
A
h
O
45o
30o
θ
C
Chapter 11 – Trigonometric (Applications)
Q24
AC =
A
x 30o
70o
35o
10
Chapter 11 – Trigonometric (Applications)
Q22
4
1
× x × 2 x sin 30o = 16
2
B
Option B
cos θ =
2
A
B
Chapter 11 – Trigonometric (Applications)
Q21
2 2 + 32 − 4 2
2(2)(3)
−1
=
4
θ
∠BAC = 70o − 35o =35o ext ∠s of ∆
∴ AC = BC = 10
AD
= sin 70o
AC
AD = 10 sin 70o
p2 + q2
∠ACB = 180o − θ
2x
cos(180o − θ ) =
C
cos θ =
−q
p2 + q2
A
q
2
p
p + q2
B
θ
q
C
Option E
4
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q25
a
2
a
P
AP
2 cos θ
a
a
a
= sin θ , BP =
BP
sin θ B θ
Q
a
a
AB = AP + PB =
+
sin θ 2 cos θ
1 
 1
= a
+

 sin θ 2 cos θ 
Option C
= cos θ , AP =
Chapter 11 – Trigonometric (Applications)
Q26
A θ
S
a
2
C
R
p−q
BC
=
sin(180o − 70o − 50o ) sin 70o
( p − q ) sin 70o
BC =
sin 60o
A
Option C
p
D
C
BC = 2 2 + 52 − 2(2)(5) cos 60o
= 19
A
2
60o
5
B
B
C
60o
A
30o
Chapter 11 – Trigonometric (Applications)
Q29
1
× AC × 6 cm × sin 30o = 15 cm 2
2
AC = 10 cm
70o 50o
Chapter 11 – Trigonometric (Applications)
Q28
sin A sin B sin C
=
=
a
b
c
a > b > c , ∠A > ∠ B > ∠ C
(b + c ) 2 − a 2 = b 2 + 2bc + c 2 − (b 2 + c 2 − 2bc cos A)
= 2bc (1 + cos A) > 0
b+c > a
Option C
B
70o
Chapter 11 – Trigonometric (Applications)
Q27
∠ B + ∠ C = ∠A
q
C
Option E
Chapter 11 – Trigonometric (Applications)
Q30
AB
1
=
sin(180o − 30o ) sin 20o
sin150o
AB =
sin 20o
1
=
2 sin 20o
Option C
B
A
1
30o
20o
C
X
Option D
5
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q32
Chapter 11 – Trigonometric (Applications)
Q31
p
p
= tan β , BD =
BD
tan β
AB
α ,α = 2
AC =
2
π −α π
β=
= −1
2
2
p
p
= tan α , CD =
CD
tan α
 1
p
p
1 
−
= p
−
BC =

tan β tan α
 tan β tan α 
C
β
A
Option E
β
B
O
Option C
Chapter 11 – Trigonometric (Applications)
Q33
AC = BC cos α = 2 cos α
BD = BC tan β = 2 tan β
The required area
1
1
= AC × BC sin α + BC × BD
2
2
= 2(sin α cos α + tan β )
1
, ∠C = 30o
2
∠A : ∠B : ∠C = 90o : (180o − 90o − 30o ) : 30Co
= 3 : 2 :1
sin ∠C =
B
D
2
α
Option E
Chapter 11 – Trigonometric (Applications)
Q34
∠A = 90o
A
β
Option C
B
C
Chapter 11 – Trigonometric (Applications)
Q35
180o − 120o
= 30o
2
AB
AD
=
sin α sin 20 o
sin 30o
1
AB =
=
o
sin 20
2sin 20o
B
1
A
Chapter 11 – Trigonometric (Applications)
Q36
α = 37 o
α=
Option B
α
The required bearing is S 37 o W
light house
α
A
20o
C
α
120o
37o
α
D
Option D
observer
6
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q37
Chapter 11 – Trigonometric (Applications)
Q38
AD
tan B
AD
DC =
tan C
AD
BD tan
tan C
= ADB =
DC tan C tan B
sin C sin B 34 1
=
= =
2
3
3 4
1
sin C =
2
1
3
2
cos C = 1 − sin 2 C = 1 − ( ) 2 =
2 A4
A
3
2
Option E
BD =
B
C
Option C
B
C
D
Chapter 11 – Trigonometric (Applications)
Q40
Chapter 11 – Trigonometric (Applications)
Q39
α =θ
With the notation in the figure
60o
θ=
= 30o , a = 18 − r
2
r
r
= sin θ ,
= sin 30o
a
18 − r
3
r = 9, r = 6
2
PR
= sin θ
x
TR
= tan θ
PR
TR
= sin θ tan θ
x
TR = x sin θ tan θ
Option D
Option D
a
r
θ
T
P
α
R
x
θ
Q
Chapter 11 – Trigonometric (Applications)
Q41
2α = 180o − 75o − 45o
α = 30o
BD
CD
=
sin 30o sin 45o
1
BD sin 30o
1
=
= 22 =
o
CD sin 45
2
2
Option B
Chapter 11 – Trigonometric (Applications)
Q42
3
4
α = 36.87o
The required angle
= 2α = 74o cor. to nearest degree
tan α =
A
D
75
o
B
45o
3
α
α
4
α
C
Option D
7
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q43
Area of I : Area of II : Area of III
1
1
1
= BC 2 sin 60o : AC 2 sin 60o : AB 2 sin 60o
2
2
2
2
2
 AB sin α   AB sin β 
2
=
 :
 : AB
C
 sin γ   sin γ 
II γ
2
2
2
α
= sin α : sin β : sin γ
β
Chapter 11 – Trigonometric (Applications)
Q44
XD
= sin θ , XD = p sin θ
p
YC
= cos(180o − φ ), YC = − q cos φ
q
I
A
CB = XD − YC = p sin θ + q cos φ D
p
θ
Option D
Chapter 11 – Trigonometric (Applications)
Q45
B
X
1
DF = CD = BC sin 30o = BC
2
BC
DB =
= 2 BC
cos 60o
DF 12 BC 1 D
sin θ =
=
=
DB 2 BC 4
B
12
A
Chapter 11 – Trigonometric (Applications)
Q46
α =θ
BC = 132 − 12 2 = 5
5
cos θ =
13
C
φ
III
Option C
Y
q
B
θ
C
α
A
O 13
Option B
Chapter 11 – Trigonometric (Applications)
Q47
F
C
Option A
∠DHG = 90o
AB = 4 k , BC = 5k , CA = 6k
∠AHG = 90o
AB 2 + CA2 − BC 2
2( AB )(CA)
(4k ) 2 + (6k ) 2 − (5k ) 2
=
9 2(4 k )(6k )
=
16
Option D
A
60o
30o
B
Chapter 11 – Trigonometric (Applications)
Q48
AB : BC : CA = 4 : 5 : 6
cos A =
E
θ
∠BEH = 90o
B
C
A
F
Option E
E
D
G
H
8
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q49
Chapter 11 – Trigonometric (Applications)
Q50
1
× AC × AB × sin 30o = 15 cm 2
2
AC ( AC − 4cm) = 60cm 2
AC 2 − 4cmAC − 60cm 2 = 0
AC = 10 cm or −C6 cm (rejected)
AB = h tan θ
AC = h tan 2θ
AC tan 2θ
=
AB
tan θ
D
θ
h θ
A
Option C
B
C
Chapter 11 – Trigonometric (Applications)
Q51
B
A
Chapter 11 – Trigonometric (Applications)
Q52
a
AB
=
o
o
sin(180 − 20 − 50 ) sin 50o
a sin 50o
a sin 50o
AB =
=
sin(180 o − 70o )
sin 70o
PQ
1
= tan 30o =
SQ
3
tan θ =
30o
Option C
o
MQ 12 PQ
1
3
=
=
=
SQ
SQ
6
2 3
P
A
M
o
30
θ
S
Option B
Q
50o
20o
Option D
B
Chapter 11 – Trigonometric (Applications)
Q53
AD
= tan θ
x
AD
= tan φ
y
AD
x tan φ
y
= =
AD
y tan θ
x
Option C
a
C
Chapter 11 – Trigonometric (Applications)
Q54
tan θ =
3
=3
1
V
A
3 cm
D
B
θ
x
φ
D
y
C
Option E
A
2 cm
C
θ
1 cm 2 cm
B
9
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q55
Chapter 11 – Trigonometric (Applications)
Q56
1
= tan 45o , a = 1
a
b
1
1
= cos 60 o , b = a =
a
2
2
The required area
1
= 10 × = 5 cm 2
2
10 2 = (2 x ) 2 + (3 x ) 2 − 2(2 x )(3 x ) cos θ
2
4 x = 100
x 2 = 25
x=5
2x
10
θ
N
AC = a 2 + a 2 = 2a
AC
a
AH =
=
2
2
sin θ
sin α
C
1
α
Option A
2
D
C
a
H
θ
A
V
a
VH = AV − AH =
2
a
1
2
sin ∠VAH =
=
a
2
2
D
1m
o
45
a
Chapter 11 – Trigonometric (Applications)
Q58
XC
1
=
sin θ sin α
Option C
B
X
Chapter 11 – Trigonometric (Applications)
Q59
A
B
Chapter 11 – Trigonometric (Applications)
Q60
α + α + β + β = 180o
3α = 360o , α = 120o
α + β = 90o
The required different
2
= 2π (1) − 3 × 12 + 12 − 2(1)(1) cos120o
AC = (5 + 5) − 6 = 8
1
Area of ∆ABC = × 6 × 8 = 24
2
= 2π − 3 3
C
β α
6
α
β
A
5
M
5
1
α α
1 α 1
5
Option C
Wall
Option B
Chapter 11 – Trigonometric (Applications)
Q57
2
10 m
b
3x
Ray of sunlight
S
60o
E
Option D
AD = XC =
Ray of sunlight
B
Option E
10
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q61
Chapter 11 – Trigonometric (Applications)
Q62
∠HAC = 90o
BD = 32 + 42 − 2(3)(4) cos(180o − 60o )
∠ABC = 90o
= 37
H
NORTH
∠HBC = 90o
A
D
A
3
60 o
Option E
B
Option E
C
4
B
Chapter 11 – Trigonometric (Applications)
Q63
Chapter 11 – Trigonometric (Applications)
Q64
α = 90o
AB = 2r
AC = AB cos 30o = 3r
C
α
o
BC = AB sin 30 = r
1
3 2
( AC )( BC ) =
r = 3A
2
2
r= 2
30o
B
Option C
Chapter 11 – Trigonometric (Applications)
Q65
YZ
XY
=
o
sin 60
sin 45o
XY sin 45o
2
=
=
o
YZ sin 60
3
2
A1 12 π ( XY2 ) 2  XY 
2
= 1 YZ 2 = 
 =
A2 2 π ( 2 )
3
 YZ 
Y
A2
A1
60
Option C
X
o
45
AB
AC
=
sin α sin 30o
DE
CD
=
sin α sin120o
AB
AC sin120o
sin α
=
DE
CD sin 30o
sin α
AB
= 3
DE
Option D
E
D
120o
α
α
C
A
30o
B
Chapter 11 – Trigonometric (Applications)
Q66
Q
P
C
o
Z
AC
BC
=
o
sin120
sin 30o
AC sin120o
=
= 3
BC sin 30o
1
1
BCh1 = ACh2
2
2
h1 AC
=
= 3
h2 BC
Option B
h1
B
120o
30o
A
h2
C
11
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q67
Chapter 11 – Trigonometric (Applications)
Q68
(35o + α ) + 75o = 180o
XO = 2a
(2b) 2 + (2c) 2
= b2 + c2
2
EX = XO 2 + EO 2
o
α = 70
The required bearing
N
N 70o E
B
A
35o 75o
B
H
2a
G
2c
O
E
0
Option B
Option D
Chapter 11 – Trigonometric (Applications)
Q69
E
F
2b
Chapter 11 – Trigonometric (Applications)
Q70
XO = 2a
h
= 3h
tan 30o
h
b=
,b = h
tan 45o
a 2 + b 2 = 3h 2 + h 2 = 302
a 2 + b 2 = 3h 2 + h 2 = 900
h = 15
a=
2
2
(2b) + (2c)
= b2 + c2
2
XO
tan θ =
A
EO
2a
=
2a
b2 + c2
D
EO =
Option E
C
X
= (2a) 2 + b 2 + c 2
α
A
D
EO =
E
C
X
B
H
θ
G
2c
O
h
45
bo
a
30
East
0
o
30 m
Option B
F
2b
P
South
Chapter 11 – Trigonometric (Applications)
Q71
Chapter 11 – Trigonometric (Applications)
Q72
10
= 10 2 cm
sin 45o
BC
10cm
=
, BC = 10 cm
o
o
o
o
sin(180 − 45 − 75 − 30 ) sin30o
AD =
AB = (10 2)2 + 102 − 2(10 2)(10) cos 75o
= 15 cm cor. to nearest cm A
D
45o
10 cm
32 + 42 − 22 7
=
2(3)(4)
8
3
B
30
Option B
cos θ =
75o
2
θ
o
Option D
4
C
12
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q73
sin θ sin135o
1
=
,sin θ =
3
3
2
cos 2 θ + sin 2 θ = 1
8
8
cos θ =
=
9
3
sin θ
1
tan θ =
=
cos θ
8
Chapter 11 – Trigonometric (Applications)
Q74
4
=2
2 2
4 + 22 + 4 2
OB =
=3
2
θ XB 2
sin =
=
2 OB 3
XB =
2 135o
θ
H
G
F
E
4
O
θ
3
C
D
Option B
A
Option A
Chapter 11 – Trigonometric (Applications)
Q75
2
B
X
Chapter 11 – Trigonometric (Applications)
Q76
O
AM = AN = OM = ON
N
2 cm
AM 2 + OM 2 = 22 , AM = 2 cm
M
A
ON
MN
2 cm C
=
, MN = 2 cm
2 cm
2 2cm
OC 2 2 cm
B
∆AMN is equilateral with sides 2 cm
The required area
1
3
= ( 2)2 sin 60o =
cm 2
2
2
Option D
Chapter 11 – Trigonometric (Applications)
Q77
B
c=6
A
30o
B
c=6
a
A
C
a
30o
C
Option C
Chapter 11 – Trigonometric (Applications)
Q78
a = 52 + 82 − 2(5)(8)( −54 )
= 153
With the notation in the figure
o
θ + φ = 180
AC = OX − r sin φ , BD = OX + r sin φ
AC + BD = 2OX
AC × BD = OX 2 − r 2 sin 2 φ
6sin 30o < a < 6
3< a < 6
r
B
r
r
a
8
Option D
C
Option A
5
A
X
13
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q79
CB = BA and BP // AX
CP = PX , AX = 2 BP
PX 1
tan θ =
=
AX 2
Chapter 11 – Trigonometric (Applications)
Q80
The required perimeter
1 
 1
= 3
+1+1+
o
tan 30o 
 tan 30
1 

= 6 1 +
o 
 tan 30 
A
θ
B
θ
1
X
P
Option E
3
2
5 + 12
2
=
3
13
30o
B
: CA
:
BC
AB = ∠A : ∠B : ∠C = 1: 2 : 3
BC : CA : AB ≠ a : b : c
E
D
F
a : b : c = sin A : sin B : sin C
3
G
Option C
A
Chapter 11 – Trigonometric (Applications)
Q82
Chapter 11 – Trigonometric (Applications)
Q81
tan θ =
1
1
C
Option C
C
θ
C
12
A
5
B
Chapter 11 – Trigonometric (Applications)
Q83
The bearing of B from A
= 180o + θ
= 180o + 75o
= 255o
Option A
Chapter 11 – Trigonometric (Applications)
Q84
θ = 180o − 120o = 60o
AC = 22 + 52 − 2(2)(5) cos 60o
A
= 19
θ
B
D
120o
C
A
75o
2
5
θ
B
Option E
Option A
14
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q85
BC = AC sin 30o = 3
PC 5
tan θ =
=
BC 3
P
Chapter 11 – Trigonometric (Applications)
Q86
3
π
= cos
AB
6
3
AB =
=2 3
cos π6
C
5
3
6
A
C
B
π
6 O
θ
30o
A
Option C
Option D
B
Chapter 11 – Trigonometric (Applications)
Q87
AC
AB
=
sin B sin C
sin B
5
2
AC = 8 ×
= 8× = 6
sin C
6
3
A
8
B
C
Option B
Chapter 11 – Trigonometric (Applications)
Q89
a 2 = b 2 + c 2 − 2bc cos150o
Chapter 11 – Trigonometric (Applications)
Q88
BD
= cos θ
AB
BD = p cos θ
BD
= tan θ
CD
p cos θ sin θ
=
CD
cos θ
p cos 2 θ
CD =
sin θ
Option E
A
D
p
θ
90o - θ
C
Chapter 11 – Trigonometric (Applications)
Q90
α = 30o
β = 45o − 30o = 15o
The bearing of B from A
a 2 = b 2 + c 2 + 3bc
= 180o − 15o = 165o
A
c
150o
θ
B
A
N
30o α β
b
O
B
Option E
a
C
B
Option D
15
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q91
BD
AD
=
o
o
sin(180 − 44 − 91 ) sin 44o
BD sin 45o
=
> 1, BD > AD
AD sin 44o
DC
AD
=
o
o
sin(91 − 47 ) sin 47o
DC sin 44o
=
< 1, DC < AD
AD sin 47o
Chapter 11 – Trigonometric (Applications)
Q92
BC
3
o
BF = BC tan 60 = 3BC
AB 1
tan θ =
=
BF 3
AB = BC tan 30o =
o
A
C
30o
60o
B
A
H
θ
91o
44o
B
Option C
D
47o
C
D
Chapter 11 – Trigonometric (Applications)
Q93
Option A
E
G
F
Chapter 11 – Trigonometric (Applications)
Q94
∠ACF = 90 o
BD = d sin θ
PD = DB tan θ
∠BDF = ∠ACE
P
PB = d sin θ tan θ
∠DEG = 60o
D
θ
A
θ
d
B
Option B
Option B
Chapter 11 – Trigonometric (Applications)
Q95
With the notation in the figure
2
2
x = 2 + 3 − 2 × 2 × 3 × cos 70
= 2.98
o
Chapter 11 – Trigonometric (Applications)
Q96
With the notation in the figure
a
= tan 35o , a = tan 35o
b
2a
tan(θ + 35o ) =
= 2 tan 35o
b
θ + 35o = 54.5o
θ o = 19.5o
a
Option B
Option B
a
b
16
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q97
Chapter 11 – Trigonometric (Applications)
Q98
With the notation in the figure
o
o
o
θ = 20 + (180 − 130 ) = 70
o
o
φ = 80 − 20 = 60
With the notation in the figure
a=2
o
o
b = 4 2 + 32 = 5
400 m
LB
=
sin60 o sin 70 o
LB =
tan θ =
θ
400 sin 70o
m
sin 60o
Option D
a 2
=
b 5
θ = 22o
a
b
θ
Option A
φ
Chapter 11 – Trigonometric (Applications)
Q99
Chapter 11 – Trigonometric (Applications)
Q100
With the notation in the figure
a
sin θ =
c
a
tan θ =
b
a a
sin θ + tan θ = +
c b
With the notation in the figure
52 + 4 2 − 7 2
cos θ =
= −0.2
2(5)( 4)
Option A
Option C
θ = 102o
Chapter 11 – Trigonometric (Applications)
Q101
Chapter 11 – Trigonometric (Applications)
Q102
With the notation in the figure
3a
tan θ =
=3
3a − a − a
1
×10 × 5 × sin ∠ABC = 18
2
sin ∠ABC = 0.72
θ = 72o
∠ABC = 46o or 134o
a
3a
Option E
Option C
a
17
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q103
Chapter 11 – Trigonometric (Applications)
Q104
With the notation in the figure
r
r
= sin α , a =
a
sin α
DC
a
r sin β
=
, DC =
sin β sin γ
sin α sin γ
With the notation in the figure
1m
= tan 45o , a = 1 m, b = a 2 + 12 = 2 m
a
1
tan θ =
, θ = 35o
2
a
θ
Option C
b
a
Chapter 11 – Trigonometric (Applications)
Q105
Option A
Chapter 11 – Trigonometric (Applications)
Q106
With the notation in the figure
With the notation in the figure
30o πa 2
Area of sector OXY = ( 2a) π ×
=
3600
3
1
a2
o
Area of ∆OXZ = (a)(2a) sin 30 =
2
2
a 2 πa 2
The ratio = :
= 3 : 2π
2 3
a = 90o − 50o = 40o
2
b = 180o − 40o − 60o = 80o
The bearing of B from A is 080o
b
2a
a
Option D
a
a
Chapter 11 – Trigonometric (Applications)
Q107
Option B
Chapter 11 – Trigonometric (Applications)
Q108
With the notation in the figure
AD
= sin 60o , AD = 4 sin 60o = 2 3 cm
4 cm
AD
= tan 30o
DC
AD 2 3
DC =
= 6 cm
tan 30o 33
With the notation in the figure
7 2 + 82 − 9 2 2
cos x o =
=
2(7)(8)
7
Option A
Option E
x o = 73.4o
18
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q109
Chapter 11 – Trigonometric (Applications)
Q110
With the notation in the figure
With the notation in the figure
SX = 6 cm, XR = 10 − 5 = 5 cm, ∠SXR = 60
o
a = 40o , b = 20 o , ∠RPS = 20o + 40o = 60o
c = 90o − 40o = 50o
PR
5m
=
sin 50o sin60o
x = 6 2 + 52 − 2(6)(5) cos 60o
= 5.57
PR =
Option B
b
a
Option B
X
Chapter 11 – Trigonometric (Applications)
Q111
θ
2
2
b = 2 2 − 12 = 3 , a = = 1
2
θ a 1
sin = =
2 b
3
c
Chapter 11 – Trigonometric (Applications)
Q112
Let X , Y be the mid - points of AB and CD respectively.
Z be the mid - point of XY
∠XVY = θ , ∠ZVY =
5 sin 50o
m
sin 60o
With the notation in the figure
a = 20o , b = 30 o , ∠BAC = 20 o + 30o = 50 o
180o − 50o
c=d =
= 65o
2
e = 180o − 30o − 65o = 85o
ab
The required bearing is S 85o W
b
d
Option C
X
Z
a
Chapter 11 – Trigonometric (Applications)
Q113
x = 12 + 12 − 2(1)(1) cos 80o (or 2 ×1sin 40o )
= 1.29
Y
Option E
c
e
Chapter 11 – Trigonometric (Applications)
Q114
AC
AB
=
sin 60o sin(180o − 60o − 75o )
AC sin 60o
6
=
=
AB sin 45o
2
Option B
Option D
19
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q115
With the notation in the figure
4 2 + 52 − 6 2 1
cos a =
=
2( 4)(5)
8
a = 82.819o
With the notation in the figure
1
tan a = , a = 26.565o
2
2
tan( x o + a) = = 1, x o + a = 45o
2
x = 45 − 26.565 = 18.4
Option B
∠ABC = 180o − 82.819o = 97 o
Option B
a
Chapter 11 – Trigonometric (Applications)
Q117
With the notation in the figure
2
Chapter 11 – Trigonometric (Applications)
Q116
a
Chapter 11 – Trigonometric (Applications)
Q118
With the notation in the figure
1
1
(10)(8) sin θ = (5)(4) sin 30o
2
2
1
sin θ =
8
θ = 7.2o
2
DM = 9 + 12 = 15 cm
DM 15 5
cos ∠EMD =
=
=
EM 36 12
∠EMD = 65o
Option C
Option A
Chapter 11 – Trigonometric (Applications)
Q119
With the notation in the figure
a
6
=
sin 50o sin(180o − 50 o − 60o )
6 sin 50o
a=
sin 70o
1
1 sin 50o sin 60o
Area = ( a)(6) sin 60o = 6 2
= 12.7 cm2
2
2
sin 70o
Chapter 11 – Trigonometric (Applications)
Q120
With the notation in the figure
a 2 = 6 2 + 52 − 2(6)(5) cos x o = 7 2 + 7 2 − 2(7)(7) cos 66o
cos x o = 0.04767
x = 87.3
a cm
Option C
Option D
a cm
20
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q121
Chapter 11 – Trigonometric (Applications)
Q122
With the notation in the figure
With the notation in the figure
k = b sin θ , h = a cos θ
BC = 2a tan 20o
BC
tan θ =
= 2 tan 20o
a
CF = h + k = a cos θ + b sin θ
θ
h cm
k cm
Option D
Chapter 11 – Trigonometric (Applications)
Q123
Option A
a
a
Chapter 11 – Trigonometric (Applications)
Q124
With the notation in the figure
With the notation in the figure
a = sin α , b = cos α ,
c = b sin β = cos α sin β ,
d = b cos β = d cos α cos β
The required volume
a = 40o , b = 90o − a = 50o , c = b = 50o
The required bearing 310o
a
1
1
= adc = sin α cos 2 α sin β cos β m3
3
3
b
a cm
c cm
b cm
Option D
c
Option E
d cm
Chapter 11 – Trigonometric (Applications)
Q125
Chapter 11 – Trigonometric (Applications)
Q126
With the notation in the figure
2
With the notation in the figure
2
x = 2 2 + 32 − 2(2)(3) cos 60o
a = (3k ) + (4k ) = 5k
sin θ =
= 7 ≈ 2.65
4k 4
=
a 5
a
4k
Option E
Option A
3k
21
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q127
With the notation in the figure
6 2 + 6 2 − (6 3 ) 2 − 1
cos θ =
=
2(6)(6)
2
o
θ = 120
arc ABC =
Chapter 11 – Trigonometric (Applications)
Q128
With the notation in the figure
θ = 180o − 54o − 66o = 60o
15 − 5
x
=
sin θ sin 54o
10 sin 54o
x=
≈ 9.34
sin 60o
120o
× (2π (6)) = 4π cm
360o
θ
θ
Option B
Option B
Chapter 11 – Trigonometric (Applications)
Q129
Chapter 11 – Trigonometric (Applications)
Q130
A
With the notation in the figure
6 3
tan θ = =
8 4
8 km
θ ≈ 36.9o
The required bearing S 36.9o E
L
With the notation in the figure
θ
6 km
B
Option E
Chapter 11 – Trigonometric (Applications)
Q131
With the notation in the figure
x
AC
=
sin 48o sin(180o − 48o − 55o )
x sin 77o
AC =
sin 48o
Option A
AB = 32 + 4 2 = 5 m
AB × CD AC × BC
3× 4
=
, CD =
= 2.4 m
2
2
5
5
25
tan θ =
=
2.4 12
Option D
Chapter 11 – Trigonometric (Applications)
Q132
1
Area of ∆AEB = (3)( 2) sin 50o = 3 sin 50o cm 2
2
Area of ABCD = 4 × Area of ∆AEB o
= 12 sin 50o = 9.2 cm2
Option C
22
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q133
Chapter 11 – Trigonometric (Applications)
Q134
22 = x 2 + (2 x ) 2 − 2( x )(2 x ) cos 70 o
x2 =
4
5 − 4 cos 70o
x=
4
≈ 1.05
5 − 4 cos 70o
With the notation in the figure
20 tan 50o − 9
20
θ ≅ 36.6o
tan θ =
θ
9m
Option B
Option C
50o
20 m
Chapter 11 – Trigonometric (Applications)
Q135
With the notation in the figure
a
b=c=
= 2a
sin 30o
FH = a 2 + a 2 = 2a
cos ∠FBH =
Chapter 11 – Trigonometric (Applications)
Q136
a
With the notation in the figure
a
c
θ = 90o − 38o = 52o
The bearing is N 52o W
b
( 2a ) 2 + ( 2a ) 2 − ( 2 a ) 2 3
=
2(2a )(2a )
4
∠FBH ≈ 41o
Option B
Chapter 11 – Trigonometric (Applications)
Q137
With the notation
in the figure
152 + 16 2 − 132
3
=
cos θ =
2(15)(16)
20
θ
Option B
Chapter 11 – Trigonometric (Applications)
Q138
∠CAF = 90o
∠DHG = 90o
∠AGC ≠ 90o
Option B
Option A
23
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q139
With the notation in the figure
PS
( y − x) cm
=
sin β sin(180 o − α − β )
Chapter 11 – Trigonometric (Applications)
Q140
With the notation in the figure
AC = 82 + 10 2 − 2(8)(10) cos 60o
= 84 ≈ 9.17
( y − x) sin β
PS =
cm
sin(α + β )
Option D
Option B
Chapter 11 – Trigonometric (Applications)
Q141
With the notation in the figure
12
a = = 6, b = 10 2 − 6 2 = 8
2
b
8 4
sin x = = =
10 10 5
Chapter 11 – Trigonometric (Applications)
Q142
With the notation in the figure
a = 12 + 12 = 2 , b = 2
2
XY = 2 2 + 2 = 6 cm
b
b
a
a
Option D
Option D
Chapter 11 – Trigonometric (Applications)
Q143
With the notation in the figure
10
tan α = = 1.125
8
10
tan β =
=1
6 2 + 82
10
10
tan γ =
=
≈ 1.17
2
2
73
3 +8
10
tan δ =
=2
32 + 4 2
Chapter 11 – Trigonometric (Applications)
Q144
With the notation in the figure
θ = 90o − 27 o = 63o
The required bearing is S 63o E
θ
Option D
α
δ
γ
β
Option D
24
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q145
Chapter 11 – Trigonometric (Applications)
Q146
With the notation in the figure
sin θ sin 30o
=
5
3
5 sin 30o 5
sin θ =
=
3
6
o
θ ≈ 56
With the notation in the figure
Option C
Option C
cos θ =
Chapter 11 – Trigonometric (Applications)
Q147
With the notation in the figure
6 2 + 6 2 − 32 7
=
2(6)(6)
8
Chapter 11 – Trigonometric (Applications)
Q148
With the notation in the figure
AB
AB
= AB, BC =
= 3 AB
tan 45o
tan 30 o
AB
1
tan ∠BCD =
=
3 AB
3
∠BCD = 30o
3
a = EB sin 30o = 2 cos 30o × sin30o =
cm
2
1
3
3
The required area = × 2 ×
=
cm 2
2
2
2
BD =
a
Option B
Option B
Chapter 11 – Trigonometric (Applications)
Q149
With the notation in the figure
a
VA = VB =
= 2a
cos 45o
Chapter 11 – Trigonometric (Applications)
Q150
With the notation in the figure
42o + 28o
∠AQP =
= 35o
2
θ = 35o − 28o = 7 o
AB = 2a cos 45o = 2a
The required bearing is N 7 o E
∆VAB is an equilateral triangle.
θ
∠AVB = 60o
Option B
a
Option A
25
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q151
Chapter 11 – Trigonometric (Applications)
Q152
With the notation in the figure
2
With the notation in the figure
The required angle is ∠BDF
2
a = 4 + (7 − 4) = 5
4
sin x =
5
a
Option C
Option B
Chapter 11 – Trigonometric (Applications)
Q153
Let M be the mid - point of AB, X is the projection of V on ABC
3 3
VM = MC = 32 − (1.5) 2 =
2 2
V
3 3 2 1 3 3 


VX = (
) − ×
 = 6 cm
2
3
2


Chapter 11 – Trigonometric (Applications)
Q154
With the notations in the figure
θ = 360o − (180o − 110o ) = 290o
The bearing of B from A is 290o
C
A
A
110o
θ B
X
M
Option C
B
Chapter 11 – Trigonometric (Applications)
Q155
With the notations in the figure
90o − 30o
∠ABD = ∠DBC =
= 30o
2
c
DC = BC tan 30o = c sin 30o × tan 30o =
2 3
Option B
Option C
Chapter 11 – Trigonometric (Applications)
Q156
With the notations in the figure
The required angle is ∠BFD
Option D
26
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q157
With the notations in the figure
x
16
=
sin(180o − 80o − 30o ) sin 80o
x=
With the notations in the figure
BD
BD
AD =
= 3BD, DC =
= BD
tan 30o
tan 45o
16 sin 70o
≈ 15
sin 80o
AD : DC = 3BD : BD = 3 : 1
Option B
Option C
Chapter 11 – Trigonometric (Applications)
Q159
With the notations in the figure
2
Chapter 11 – Trigonometric (Applications)
Q158
2
2
12 + 5 = 144 + 25 = 169 = 13 , α = 90
5
tan θ =
12
Chapter 11 – Trigonometric (Applications)
Q160
With the notations in the figure
o
a = 130o − 60 o = 70o , b = 120o
BD (150 − 80) cm
=
sin b
sina
BD =
70 sin 120o
cm ≈ 65 cm
sin 70o
α
α
Option A
Chapter 11 – Trigonometric (Applications)
Q161
Option D
b
Chapter 11 – Trigonometric (Applications)
Q162
HF = BF 2 + 2 2 , GE = AE 2 + 32
With the notations in the figure
BF = AE < HF < GE
θ = 40o
CF DE CF DE
=
>
>
BF AE HF GE
The angle of depression of Q from P is 40o
tan b = tan a > tan c > tan d
P
d <c<a =b
θ
40o
Q
Option D
Option A
27
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q163
With the notations in the figure
AB 3
tan θ =
=
BC 2
θ ≈ 56o
Chapter 11 – Trigonometric (Applications)
Q164
With the notations in the figure
1
AC
4
Area of ∆ACD = 4 × Area of ∆ADE = 4 cm 2
AC cos 60o = AD, AE = AD cos 60o =
Area of ∆ABC = Area of ∆ACD = 4 cm 2
Option D
Option B
Chapter 11 – Trigonometric (Applications)
Q165
With the notations in the figure
Chapter 11 – Trigonometric (Applications)
Q166
With the notations in the figure
7 sin(330o + θ ) = 0
AD = a − b
= BC sin x − CD cos y
330o + θ = 360o
θ = 30o
a
Option B
Option C
b
Chapter 11 – Trigonometric (Applications)
Q167
With the notations in the figure
12
XN = = 6 cm
2
MN = 82 + 152 = 17 cm
tan θ =
Chapter 11 – Trigonometric (Applications)
Q168
With the notations in the figure
CD DE
∆CDE ~ ∆BAE ,
=
AB AE
o
a = 90
XN
6
=
MN 17
N
CD
= cosθ
AB
a
θ
Option B
Option B
28
10/12/2015
Chapter 11 – Trigonometric (Applications)
Q169
Chapter 11 – Trigonometric (Applications)
Q170
With the notations in the figure
With the notations in the figure
AD = AC sin β
AD
AC sin β
BD =
=
tan α
tan α
AC tan α
=
BD sin β
The required angle is ∠AED
Option B
Option D
Chapter 11 – Trigonometric (Applications)
Q171
With the notations in the figure
2
 3 
( 2a ) 2 − a 2
3
7
AD =
=
a, BD = a 2 + 
a  =
a
2
2
2
2


o
sin θ sin 30
=
AD
BD
3
a× 1
21
sin θ = 2 7 2 =
30o
14
a
2
Chapter 11 – Trigonometric (Applications)
Q171
With the notations in the figure
MC = NC = a 2 + ( 2a) 2 = 5a
MN = ( 2a) 2 + (2a) 2 = 2 2a
cosθ =
MN
2
MC
=
2
10
=
5
5
a
2a
2a
Option D
Option D
a
2a
a
29