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Sine Ratio
Introduction to Trigonometric Ratios
The figure below shows a right-angled triangle ABC,
where B = and C = 90.
A
hypotenuse
θ
B
adjacent side
of
opposite
side of
C
AB is called the hypotenuse;
BC is called the adjacent side of ;
AC is called the opposite side of .
In
fact, thethe
size
of has certain
relationship
Consider
right-angled
PQR
PQ PQ among
QR
between
ratios
below. Isthe
there
any relationship
and
,
PR .
PR QR
, PQ,ratios
QR and
These
arePR?
known as trigonometric
ratios.
P
θ
Q
R
2 + QR2 = PR2…
IHow
onlydoes
knowthe
thatsize
PQof
relate to the
sides of the triangle?
Consider
threeWhat
right-angled
Completethe
thefollowing
table below.
do you
triangles.
observe?
4
2
1
A
Triangle
opposite side
hypotenuse
2
B
30
3
6
C
30
30
A
B
C
1
2
1
2
=
2
4
1
3
=
2
6
For a right-angled triangle with a given acute
angle , opposite side is a constant.
hypotenuse
Concept of Sine Ratio
The sine ratio of an acute angle is defined as below:
opposite side
sin θ
hypotenuse
hypotenuse
θ
opposite
side of
For a right-angled triangle with a given acute angle ,
the sine ratio of is a constant.
For example,
2
30
sin 30 =
6
4
1
2
30
1
2
3
=
=
2
4
6
30
3
In
ABC, C 90, AB 13 and AC 5.
A
13
B
AC
AC
sin
B
sin B
AB
AB
5 5
13 13
5
C
AC is the opposite side of B,
and AB is the hypotenuse.
Follow-up question 1
In the following figures, find sin θ.
(a)
(b)
15
12
17
θ
θ
20
16
8
Solution
15
(a) sin θ
17
(b)
16
sin θ
20
4
5
Example 1
In △PQR, ∠P 90, PQ 6, QR 10
and RP 8. Find the values of
(a) sin ∠Q,
(b) sin ∠R.
(Give your answers in fractions.)
Solution
PR
(a) sin Q
QR
8
10
4
5
PQ
(b) sin R
QR
6
10
3
5
Finding Sine Ratio Using Calculators
Find sin for a given angle
1. Make sure that the calculator is set in degree mode.
Degree mode is usually denoted by the key DEG or D
on calculators.
2. Use the key sin on a calculator to find the value of sin .
For example,
the value of sin 30 can be obtained by keying:
sin 30 EXE
The answer is 0.5.
Follow-up question 2
By using a calculator, find the values of the following
expressions correct to 4 significant figures.
(a) sin 43 – sin 28
(b) 2 sin 11
Solution
(a) sin 43 – sin 28
= 0.2125
(b) 2 sin 11
= 0.3816
sin 43 = 0.681 99…, sin 28 = 0.469 47…
(cor. to 4 d.p.)
sin 11 = 0.190 80…
(cor. to 4 d.p.)
Example 2
By using a calculator, find the values of the following
expressions correct to 4 decimal places.
(a)
sin 66
Solution
(a)
(b) sin 32.48
Keying sequence
Display
sin 66 EXE
0.913545457
sin 66 0.9135 (cor. to 4 d.p.)
(b)
Keying sequence
Display
sin 32.48 EXE
0.537005176
sin 32.48 0.5370 (cor. to 4 d.p.)
Example 3
(a)
By using a calculator, find the value of sin 34 + sin 26
(b)
sin 60 correct to 3 significant figures.
From the result obtained in (a), is sin 34 + sin 26 equal
to sin (34 + 26)?
Solution
(a)
Keying sequence
Display
sin 34 + sin 26 –sin 60 EXE
0.131538646
sin 34 sin 26 sin 60 0.132 (cor. to 3 sig. fig.)
(b) ∵ sin 34 sin 26 sin 60 0
sin 34 sin 26 sin 60
sin 34 sin 26 sin (34 26)
Find for a given value of sin
In degree mode, use the keys SHIFT and sin to find
the corresponding acute angle .
For example,
given that sin = 0.5, can be obtained by keying:
SHIFT
sin 0.5 EXE
The answer is 30, i.e. = 30.
Follow-up question 3
Find the acute angle in each of the following using a
calculator. (Give your answers correct to 3 significant figures.)
(a) sin = 0.22
(b) sin = sin 68 – sin 40
Solution
(a) sin = 0.22
= 12.7
(cor. to 3 sig. fig.)
(b) sin = sin 68 – sin 40
sin 68 = 0.927 18…, sin 40 = 0.642 78…
= 0.2844…
= 16.5 (cor. to 3 sig. fig.)
Example 4
Find the acute angles in the following using a calculator.
(a) sin 0.62, correct to the nearest degree.
1
sin 35, correct to the nearest 0.1.
5
(b)
sin
(c)
7 sin 3, correct to 3 significant figures.
Solution
(a)
Keying sequence
Display
SHIFT sin 0.62 EXE
38.31613447
sin 0.62
38 (cor. to the nearest degree)
(b)
Keying sequence
Display
SHIFT sin ( 1 5 sin 35 ) EXE
6.587203533
1
sin sin 35
5
6.6 (cor. to the nearest 0.1)
(c)
7 sin 3
3
sin
7
Keying sequence
Display
SHIFT sin ( 3 7 ) EXE
25.37693352
3
sin
7
25.4 (cor. to 3 sig. fig.)
Using Sine Ratio to Find Unknowns in
Right-Angled Triangles
We can use the sine ratio to solve problems involving
right-angled triangles.
In
ABC, C = 90, B = 55 and AB = 8 m.
A
Find AC correct to 2 decimal places.
AC
sin B
AB
AC
sin 55
8m
AC 8 sin 55 m
to 2tod.p.)
6.55 m (cor.
(cor.
2 d.p.)
8m
55
B
C
In
PQR, R = 90, PQ = 9 m and PR = 7 m.
P
Find Q correct to 2 decimal places.
AC
sin B
AC
PR
sin B
Q AB
AB
PQ
AC
sin 55
7
m
AC
8
m
sin 55
Q
m 55 m
AC 89 sin
m
AC
sin 55
m to 2 d.p.)
Q
8
51.06
(cor.
6.55 m
(cor. to 2 d.p.)
7m
9m
R
Q
Follow-up question 4
In
C
ABC, B 75, C 90 and AC 7 cm.
Find AB correct to 2 decimal places.
Solution
AC
AC
sin
B
sin B
AB
AB
m
77cm
cm
sin
sin75
75
AB
AB
AB
77
AB
cm
m
AB
cm
sin
sin75
75
7.25
(cor.
to2d.p.)
2d.p.)
d.p.)
m
(cor.
7.25cm
cm (cor.
toto2
7 cm
A
75
B
Example 5
In △ABC, ∠B 90, ∠C 42 and
AC 5 cm. Find the length of AB correct
to 1 decimal place.
Solution
∵
AB
sin C
AC
∴
sin 42
AB
5 cm
AB 5 sin 42 cm
3.3 cm (cor. to 1 d.p.)
Example 6
In △ABC, ∠B 38, ∠C 90 and
AC 15 cm. Find the length of AB correct
to 1 decimal place.
Solution
∵
∴
AC
sin B
AB
15 cm
sin 38
AB
15
AB
cm
sin 38
24.4 cm (cor. to 1 d.p.)
Example 7
In △ABC, ∠C 90, AB 17 cm and
BC 9 cm. Find ∠A correct to the
nearest degree.
Solution
∵
∴
BC
sin A
AB
9 cm
17 cm
A 32 (cor. to the nearest degree)
Cosine Ratio
Concept of Cosine Ratio
The cosine ratio of an acute angle is defined as below:
adjacent side
cos θ
hypotenuse
hypotenuse
θ
adjacent side of
For a right-angled triangle with a given acute angle ,
the cosine ratio of is a constant.
For example,
6
4
2
60
1
cos 60 =
60
60
2
1
2
3
=
=
2
4
6
3
In
ABC, C = 90, AB = 5.2 and AC = 2.
A
5.2
B
AC
AC
AC
AB
AB
AB 2
2
2 5.2
5.2
5.2 5
5
5
13
13
13
cos A
cos
cos
A
A
2
C
AC is the adjacent side of A,
and AB is the hypotenuse.
Follow-up question 5
In the following figures, find cos θ.
(a)
(b)
6
θ 3
5
8
θ
10
4
Solution
(a)
cos θ
3
5
(b)
8
10
4
5
cos θ
Example 8
In △PQR, ∠P 90, PQ 20, PR 21
and RQ 29. Find the values of
(a) cos ∠Q,
(b) cos ∠R.
(Give your answers in fractions.)
Solution
PQ
(a) cos Q
QR
20
29
PR
QR
21
29
(b) cos R
Finding Cosine Ratio Using Calculators
Find cos for a given angle
In degree mode, use the key cos to find the value of cos .
For example,
the value of cos 30 can be obtained by keying:
cos 30 EXE
The answer is 0.8660…
Follow-up question 6
By using a calculator, find the values of the following
expressions correct to 3 significant figures.
cos 40 cos 75
(a) 5 cos 29
(b)
2
Solution
(a) 5 cos 29
= 4.37
(b)
cos 29 = 0.874 61…
(cor. to 3 sig. fig.)
cos 40 cos 75 cos 40 = 0.766 04…, cos 75 = 0.258 81…
2
= 0.512 (cor. to 3 sig. fig.)
Example 9
By using a calculator, find the values of the following
expressions correct to 4 decimal places.
(a)
cos 12.3
(c)
cos 10
cos 72
5
(b)
7
cos 81
5
Solution
(a)
Keying sequence
Display
cos 12.3 EXE
0.977045574
cos 12.3 0.9770 (cor. to 4 d.p.)
(b)
Keying sequence
Display
( 7 5 ) cos 81 EXE
0.219008251
7
cos 81 0.2190 (cor. to 4 d.p.)
5
(c)
Keying sequence
Display
cos 72 – cos 10 5 EXE
0.112055443
cos 10
cos 72
0.1121 (cor. to 4 d.p.)
5
Find from a given value of cos
In degree mode, use the keys SHIFT and cos to
find the corresponding acute angle .
For example,
given that cos = 0.5, can be obtained by keying
SHIFT
cos 0.5 EXE
The answer is 60, i.e. = 60.
Follow-up question 7
Find the acute angle in each of the following using a
calculator. (Give your answers correct to 3 significant figures.)
(a) cos θ 0.474
cos 24
(b) cos θ
2
Solution
(a)
cos
θ
0.474
1.
cos
θ
0.474
1. cos θ 0.474
61.7
(cor.
sig. fig.) degree)
θθ 62
(cor.
to to
the3 nearest
cos 24
24
cos
2.
cos
θ
2. cos θ
cos 24 = 0.913 54…
(b)
2
2
62.82
(cor.
4 sig.
fig.)
θθ 62.8
(cor.
to to
3 sig.
fig.)
Example 10
Find the acute angles in the following using a calculator.
(a)
(b)
cos 0.583, correct to the nearest degree.
cos 2 cos 75, correct to the nearest 0.1.
(c)
12 cos 5, correct to 3 significant figures.
Solution
(a)
Keying sequence
Display
SHIFT cos 0.583 EXE
54.33817552
cos 0.583
54 (cor. to the nearest degree)
(b)
Keying sequence
Display
SHIFT cos ( 2 cos 75 ) EXE
58.8260478
cos 2 cos 75
58.8 (cor. to the nearest 0.1)
(c) 12 cos 5
cos
5
12
Keying sequence
Display
SHIFT cos ( 5 12 ) EXE
65.37568165
5
cos
12
65.4 (cor. to 3 sig. fig.)
Using Cosine Ratio to Find Unknowns in
Right-Angled Triangles
We can use the cosine ratio to solve problems involving
right-angled triangles.
In ABC, C = 90, B = 55 and AB = 8 m.
Find BC correct to 2 decimal places.
BC
BC
8m
cos
cos
B
B
BC
AB
cos B AB
BC
AB
55
cos 55
40 BC
B
9
m
8
cos 40
9 cos
m 40 m
BC 9
BC
98 cos
cos
m
BC
4055
m
6.89
m
(cor.
to
2
d.p.)
6.89
m
(cor.
to
2
4.59 m (cor. to 2 d.p.)d.p.)
A
C
In PQR, R = 90, PQ = 9 m and QR = 7 m.
Find Q correct to 2 decimal places.
P
9m
BC
cos B QR
cos Q AB
BC
cos B PQ
BC
cos 40 AB
BC
7m
9
cos
40Q
cos
m
m
9cos
m 40 m
BC 99
BC 9 cos 40 m
Q 38.94 (cor. to 2 d.p.)
(cor.
6.89
6.89 m
m
(cor. to
to 2
2 d.p.)
d.p.)
Q
7m R
Follow-up question 8
In
ABC, C 90, AB 4 cm and BC 3.5 cm.
B
Find B correct to 2 decimal places.
3.5 cm
C
4 cm
Solution
BC
BC
cos
BB BC
cos
cos B AB
AB
AB
BC
BC
cos35
35 3.5 cm
cos
cos B 44 cm
cm
4 cm
BC 44 cos
cos 35
35 cm
cm
BC
BC 4 cos 35 cm
2 d.p.)
B 3.28
28.96cm
(cor.(cor.
to 2to
d.p.)
3.28 cm
(cor. to 2 d.p.)
A
Example 11
In △DEF, ∠D 90, ∠E = 62 and
EF = 8 cm. Find the length of DE
correct to 1 decimal place.
Example 12
In △PQR, ∠P 36, ∠Q 90 and
PQ 10 cm. Find the length of PR
correct to 1 decimal place.
Example 13
In △PQR, ∠R 90, PQ = 22 cm
and QR =18 cm. Find ∠Q correct to
the nearest 0.01.
Example 11
In △DEF, ∠D 90, ∠E = 62 and
EF = 8 cm. Find the length of DE
correct to 1 decimal place.
Solution
∵
DE
cos E
EF
∴
cos 62
DE
8 cm
DE 8 cos 62 cm
3.8 cm (cor. to 1 d.p.)
Example 12
In △PQR, ∠P 36, ∠Q 90 and
PQ 10 cm. Find the length of PR
correct to 1 decimal place.
Solution
∵
∴
PQ
cos P
PR
10 cm
cos 36
PR
10
PR
cm
cos 36
12.4 cm (cor. to 1 d.p.)
Example 13
In △PQR, ∠R 90, PQ = 22 cm
and QR =18 cm. Find ∠Q correct to
the nearest 0.01.
Solution
∵
∴
QR
cos Q
PQ
18 cm
22 cm
Q 35.10 (cor. to the nearest 0.01)
Tangent Ratio
Concept of Tangent Ratio
The tangent ratio of an acute angle is defined as below:
opposite side
tan θ
adjacent side
θ
adjacent side of
opposite
side of
For a right-angled triangle with a given acute angle ,
the tangent ratio of is a constant.
For example,
1
45
1
tan 45 =
3
2
45
2
1
2
3
=
=
1
2
3
45
3
In
ABC, C = 90, AC = 2.4 and BC = 3.2.
A
2.4
B
AC
tan B
BC
2.4
2.4
3.2
3.2
33
44
3.2
C
AC is the adjacent side of A,
and BC is the opposite side of A.
Follow-up question 9
In the following figures, find tan θ.
(a)
(b)
5
5
θ
θ
12
13
4
Solution
(a)
12
tan θ
5
(b)
3
tan θ
4
3
Example 14
In △PQR, ∠R 90, PQ 37,
PR 12 and RQ 35. Find the values
of
(a) tan∠P,
(b) tan∠Q.
(Give your answers in fractions.)
Solution
(a)
QR
tan P
PR
35
12
PR
(b) tan Q
QR
12
35
Finding Tangent Ratio Using Calculators
Find tan for a given angle
In degree mode, use the key tan to find the value of tan .
For example,
the value of tan 45 can be obtained by keying:
tan 45 EXE
The answer is 1.
Follow-up question 10
By using a calculator, find the values of the following
expressions correct to 4 significant figures.
tan 43 2
(a) 7 tan 51
(b)
tan 57
Solution
(a)
(b)
tan 51 = 1.234 89…
7 tan 51
8.644 (cor. to 4 sig. fig.)
tan 43 2
2.
(cor. to 4 sig. fig.)
tan 43tan
257 1.904
tan 43 = 0.932 51…, tan 57 = 1.539 86…
tan 57
1.904 (cor. to 4 sig. fig.)
Example 15
By using a calculator, find the values of the following
expressions correct to 4 significant figures.
(a)
tan 28.26
(b)
tan 65.32 tan 46.15
Solution
(a)
tan 28.26 0.5375
(cor. to 4 sig. fig.)
(b) tan 65.32 tan 46.15 2.265
(cor. to 4 sig. fig.)
Find for a given value of tan
In degree mode, use the keys SHIFT and tan to find
the corresponding acute angle .
For example,
given that tan = 1, can be obtained by keying:
SHIFT
tan 1 EXE
The answer is 45, i.e. = 45.
Follow-up question 11
Find the acute angle in each of the following using a
calculator. (Give your answers correct to 3 significant figures.)
(a)
tan θ 2.77
(b)
tan θ 3 tan 20
Solution
(a)
1. tan θ 2.77
1. tan θθ 70
2.77
(cor.
to to
the3 nearest
70.1
(cor.
sig. fig.) degree)
θ
20
(cor.
the
nearest
70.1
(cor.
3 sig.
fig.) degree)
2. tan θ
70
3
tan
toto
tan to
204=sig.
0.363fig.)
97…
(b)
tan 20
2. tan θθ 3
47.52
(cor.
θ 47.52
(cor.
4 sig.
47.5 (cor.
to 3tosig.
fig.)fig.)
Example 16
Find the acute angles in the following using a calculator.
(a)
(b)
tan = 6.54, correct to the nearest degree.
tan 2 tan 62 + 1, correct to 2 decimal places.
(c)
9 tan 2, correct to 4 significant figures.
Solution
(a)
tan 6.54
81 (cor. to the nearest degree)
(b) tan 2 tan 62 1
78.14 (cor. to 2 d.p.)
(c)
9 tan 2
12.53 (cor. to 4 sig. fig.)
Using Tangent Ratio to Find Unknowns in
Right-Angled Triangles
We can use the tangent ratio to solve problems
involving right-angled triangles.
In
ABC, B = 50, C = 90 and BC = 12 m.
Find AC correct to 2 decimal places.
AC
AC
AC
tan
B
tan
B
tan B BC
BC
BC
AC
AC
AC
tan
50
tan
50
tan 50 12
m
12
12 m
m
AC
12 tan
50 m
AC
AC 12
12 tan
tan 50
50 m
m
14.30
m
(cor.
to
22 d.p.)
(cor.
tod.p.)
d.p.)
14.30
14.30 m
m (cor.
to 2
A
B
50
12 m
C
In PQR, R = 90, PR = 16 m and
QR = 13 m. Find Q correct to 2
decimal places.
AC
tan B AC
PR
tan
BC
tan
Q
B QR
BC
AC
tan 50 16
ACm
tan
12 m
tan 50
Q 12
m
AC 13
12 m
tan 50 m
AC
tan
50
m
AC
12
12
tan
50
m to 2 d.p.)
50.91 (cor.
(cor.
14.30
14.30 m
m
(cor. to
to 2
2 d.p.)
d.p.)
P
Q
16 m
13 m
R
Follow-up question 12
A
In ABC, C 90, BC 2.4 cm and AC 7 cm.
AC = 7 cm. Find B correct to 2 decimal
places.
7 cm
C
2.4 cm
AC
Solution
tan B
BC
AC
tan B 7 cm
tan 51 BC
BC
77cm
cm
tan 51
B
7
BCcm cm
BC 2.4
tan751
BC
B 71.08 cm
(cor. to 2 d.p.)
tan 51
5.67 cm
(cor. to 2 d.p.)
B
Example 17
In △PQR, ∠P = 65.2, ∠Q = 90 and
PQ = 4 cm. Find the length of QR
correct to 3 significant figures.
Example 18
In △PQR, ∠P = 90, ∠R = 42.6 and
PQ = 6.5 cm. Find the length of PR correct
to 3 significant figures.
Example 19
In △PQR, ∠Q = 90, PQ = 15 cm and
QR = 12 cm. Find ∠P and ∠R correct to
the nearest 0.1.
Example 17
In △PQR, ∠P = 65.2, ∠Q = 90 and
PQ = 4 cm. Find the length of QR
correct to 3 significant figures.
Solution
∵
∴
QR
tan P
PQ
QR
tan 65.2
4 cm
QR 4 tan 65.2 cm
8.66 cm (cor. to 3 sig. fig.)
Example 18
In △PQR, ∠P = 90, ∠R = 42.6 and
PQ = 6.5 cm. Find the length of PR correct
to 3 significant figures.
Solution
∵
∴
PQ
tan R
PR
6.5 cm
tan 42.6
PR
6. 5
PR
cm
tan 42.6
7.07 cm (cor. to 3 sig. fig.)
Example 19
In △PQR, ∠Q = 90, PQ = 15 cm and
QR = 12 cm. Find ∠P and ∠R correct to
the nearest 0.1.
Solution
QR
PQ
12 cm
15 cm
∵
tan P
∴
P 38.7 (cor. to the nearest 0.1)
∵
tan R
∴
R 51.3 (cor. to the nearest 0.1)
PQ
QR
15 cm
12 cm
Simple Applications of
Trigonometric Ratios
Solving Problems Involving Plane Figures
For plane figures involving right-angled triangles, we can
use sine, cosine or tangent ratio to find the length of an
unknown side or the size of an unknown angle.
Can you find the length of BC in
the figure? Give your answer
correct to 4 significant figures.
A
8 cm
B
40
35
D
C
In ABD
ABD,
In
,,
In
ABD
AD
AD
AD sin 35
sin35
sin35
AB
AB
AB
AD
sin35
AD AB
AB sin35
sin
35
88
8sin35
sin
cm
cm
sin35
cm
sin 35
35cm
cm
BD
BD
BD
cos
cos35
35
cos35
AB
AB
AB
BD
AB
ABcos35
cos
BD
BD
AB
cos35
cos 35
35
88
8cos35
cos
cm
......(1)
(1)
cos35
cm
......
(1)
cos 35
35cm
cm......
cm
......
(1)
In ACD
ACD,,,
In
In
ACD
AD
AD
AD
tan40
tan 40
tan40
CD
CD
CD
AD
AD
AD
CD
CD
CD
tan 40
AB
AB
BD
BDAB
ABcos35
cos35
88cos35
cos35cm
cm......
...... (1)
(1) A
InInIn ACD,
ACD,,
ACD
8 cm
AD
AD
AD
tan
40
tan40
tan40
35
B
CD
CD
CD
D
AD
AD
CD
CD
tan 40
tan40
sin35
88sin
35
cm ...... (2)
cm
tan40
tan
40
40
C
BC
BCBD
BDDC
DC
sin35
35
88sin
cos35
35
cm
88cos
cm
tan40
40
tan
12.02cm
cm (cor.
12.02
(cor.toto44sig.
sig.fig.)
fig.)
Follow-up question 13
A
In the figure, AD is the height
,
In ofABDABC
,
5 cm
60
AB 5 cm, BAD 60 and
ACB
45
.
AD
cos 60
45
AB
B
Find AC.
D
AD
AB
cos
60
(Give your answer correct to 3 significant figures.)
5 cos 60 cm
Solution
In ACD,
2.5 cm
In
,
In ABD
ABD,
AD
sin 45
AD
AC
cos 60
AB
AD
AC
AD AB cos 60
sin 45
5 cos 60 cm
2.5 cm
2.5 cm
sin 45
AD
3.54 cm (cor.
(cor.
tosig.
3 sig.
fig.)
to
3
fig.)
sin 45
AC
C
Example 20
In the figure, BD is the height of △ABC,
BC = 8 m, ∠BAC = 70 and ∠CBD = 60.
Find AC. (Give your answer correct to 2
decimal places.)
Solution
In △BCD,
BD
cos 60
BC
BD BC cos 60
8 cos 60 m
4m
DC
sin 60
BC
DC BC sin 60
8 sin 60 m (1)
In △ABD,
BD
tan 70
AD
BD
AD
tan 70
4
m (2)
tan 70
∴
AC AD DC
4
8 sin 60 m
tan 70
8.38 m (cor. to 2 d.p.)
from (1) and (2)
Example 21
In the figure, PQRS is a trapezium with
∠R = ∠S = 90, ∠Q = 52, PQ = 6 cm
and PS = 5 cm. Find the area of PQRS
correct to 2 decimal places.
Solution
Draw PT QR as shown in the figure.
In △PQT,
PT
PQ
PT PQ sin 52
sin 52
6 sin 52 cm
QT
cos 52
PQ
QT PQ cos 52
6 cos 52 cm
∴
1
Area of PQRS ( PS QR ) PT
2
1
2
(5 6 cos 52 5) 6 sin 52 cm
2
32.37 cm 2 (cor. to 2 d.p.)
Example 22
The figure as shown is formed by a
right-angled triangle PQR and two
hemispheres with diameters PQ and RQ.
∠PQR = 90, ∠PRQ = 25 and
PR = 16 cm. Find the perimeter of the
figure correct to the nearest cm.
Solution
In △PQR,
PQ
sin 25
PR
∴ PQ PR sin 25
16 sin 25 cm
QR
cos 25
PR
∴ QR PR cos 25
16 cos 25 cm
1
1
Perimeter of the figure PQ QR 16 cm
2
2
1
1
16 sin 25 16 cos 25 16 cm
2
2
49 cm (cor. to the nearest cm)
Solving Real-life Problems
We can also use trigonometric ratios to solve real-life problems
involving right-angled triangles.
Let’s study the example on the next page.
A rectangular advertising board is
fixed to a vertical wall and is
supported by two straight cable
wires AB and AC, as shown in the
figure. It is known that ABD = 30, B
ACD = 60 and CD = 2 m.
Find AC and AB.
(Give your answers correct to
3 significant figures if necessary.)
A
60
30
C
D
2m
A
In
ACD,
In
In ACD
ACD,,
In
ACD,
CD
CD
CD cos
60
cos
cos
60
CD
AC
AC
AC cos 60
AC
CD
CD
CD
AC
AC
AC cos
CD
cos
AC cos 60
60
cos2260
2 m
m
m
2
cos
60
cos
60
cos 60 m
cos
60
44
m
4m
m
4m
AD
AD
AD tan
60
tan
tan
60
AD
CD
CD
CD tan 60
CD CD tan 60
AD
AD
AD CD
CD tan 60
AD
2CD
tan
60
tan
60
m
2
tan
60
m
2 tan 60 m
2 tan 60 m
B
30
60
C 2m D
CD
CD
CD cos
cos
60
60
cos
60
AC
AC
AC
CD
CD
CD
AC
AC
AC
cos 60
cos
cos 60
60
22
2 m
cos 60 m
m
cos
60
cos 60 B
m
44
4m
m
AD
AD
AD tan
tan
60
60
tan
60
CD
CD
CD
AD
CD
tan
60
AD
AD
CD
CD tan
tan 60
60
tan
60
m
22
2 tan
tan 60
60m
m
In
ABD,
AD
sin 30
AB
AD
AB
sin 30
A
In ABD,
60
30AD
Csin2 30
m D
AB
AD
AB
sin 30
2 tan 60
m
sin 30
6.93 m
(cor. to 3 sig. fig.)
Follow-up question 14
A rectangular advertising board is fixed to
a vertical wall and is supported by two
straight cable wires AB and AC, as
shown in the figure. It is known that
B
ACB 50, AC 1m and BC 1.7 m.
Find ABC.
(Give your answer correct to 3
significant figures.)
A
1m
50
D
1.7 m
C
Follow-up question 14 (cont’d)
In
In
ADC
A,,
ADC
AD
AD
sin
sin50
50 1 m
AC
AC
C
AD
50
AD AC
AC sin
sin50
50
B
D 50 m
11sin
sin
50 m
1.7 m
Solution
sin
sin50
50m
m
DC
DC
In ADC,
In
ADC,
cos
cos 50
50
AC
AC
AD
AD
sin
sin50
50
DC
DC AC
AC cos
cos 50
50
AC
AC
11cos
cos 50
50m
m
AD
AD AC
AC sin
sin50
50
cos 50
50m
m
cos
11sin
sin50
50 m
m
sin
sin50
50m
m
DC
DC
cos
cos50
50
AC
AC
BD BC DC
(1.7 cos 50) m
Follow-up question 14 (cont’d)
A
1m
B
Solution
In
ABD,,
In
ABD
In ABD,
50
D
1.7 m
C
AD
AD
tan
ABD
ABD
tan
ABD
tan
BD
BD
BD
sin50
50
sin
50
sin
1.7 cos 50
1.7 cos
cos 50
50
1.7
ABD
ABD 35.9
35.9 (cor.(cor.
(cor.
to
sig.
fig.)
ABD
35.9
to 3to
sig.
fig.)fig.)
33 sig.
i.e.
ABC
ABC 35.9
35.9
i.e.
ABC
35.9
i.e.
Example 23
There is a fish pond between A and B. A man
wants to go from A to B. He walks for 60 m
from A to C, then turns 75 clockwisely and
walks for 42 m from C to B. If ∠BAC = 30,
find AB. (Give your answer correct to 1
decimal place.)
Solution
Draw CD AB as shown in the figure.
In △ACD,
AD
cos 30
AC
∴ AD AC cos 30
60 sin 30 m (1)
ACD 180 90 30 ( sum of △)
60
BCD 180 75 60 (adj. s on st. line)
45
In △BCD,
DB
BC
DB BC sin BCD
42 sin 45 m (2)
sin BCD
∴
∴
AB AD DB
(60 cos 30 42 sin 45) m
81.7 m (cor. to 1 d.p.)
Example 24
Find the values of the following
trigonometric ratios using the quarter of
the unit circle as shown. (Give your
answers correct to 1 decimal place.)
(a)
(c)
sin 65
cos 46
(b) sin 12
(d) cos 84
Solution
(a) Construct line segment OP such that
OP makes an angle 65 with the
positive x-axis.
sin 65 y -coordinate of P
0.9
(b) Construct line segment OQ such that
OQ makes an angle 37 with the
positive x-axis.
sin 12 y -coordinate of Q
0.2
(c) Construct line segment OR such that
OR makes an angle 46 with the
positive x-axis.
cos 46 x-coordinate of R
0.7
(d) Construct line segment OS such that
OS makes an angle 84 with the
positive x-axis.
cos 84 x-coordinate of S
0.1
Trigonometric Ratios of Special Angles
In
general,
the the
values
shown
In
fact,
the
exact
values
of
the
Can
you
find
out
value
of
the calculator
trigonometric
ratios screen
of someare
sinon
60°?
approximations
only.
special
angles such
as 30°, 45°
and 60° can be deduced from
the properties of triangles.
With a calculator, I can evaluate
sin 60° = 0.866 025 403...
First, let’s review on the trigonometric
ratios and the Pythagoras’ theorem.
Consider right-angled triangle ABC,
we have
1. sin B b
c
c
a
cos B
B
c
a
b
tan B
a
2. By Pythagoras’ theorem,
c 2 a2 b2
A
b
C
Using the above knowledge and
considering the following triangles,
we can find the exact values of the
trigonometric ratios of 30°, 45° and
60°.
R
C
60°
45°
A
45°
1
B
2
2
1
P
60°
60°
2
Q
Trigonometric Ratios of 45°
C
First,
find
the
exact
values
Consider
isosceles
right-angled
can
apply of
Sincelet’s
= 90
Bthe
°, we
the
trigonometric
ratios
45AC.
°.
triangle
ABC theorem
on the
right.
Pythagoras’
toof
find
45°
1
A
45°
1
2
2
AC (AB
(Pyth. theorem)
1 )2 ______
1 ) (BC
We have
1
2
BC
or
sin 45° =
2
2
AC
cos 45° =
AB
1
2
or
2
AC
2
BC
tan 45° = 1
AB
B
Trigonometric Ratios of 60° and 30°
First
construct
asides,
perpendicular
line
Now,
Since
we
PRS
can
and
find
QRS
the
trigonometric
aretrigonometric
two congruent
Now,
let’s
try
to
find
the
We
have
found
the
exact
values
of from R
PS
=
QS
(corr.
s).
Consider
the
triangle
PQR.
First,
find
and PRS.
PQR
isfind
anPS
equilateral
triangle.
Then,
RS.
and
meet
PQ
at
S.
ratios
right-angled
of
60
and
triangles,
30
.
ratios
of
special
angles
and 30°.
°
° of60°
the trigonometric
ratios
45°.
Consider △PRS. PS ___
30°
1 and PRS ____
2
2
RS (RP
3 (Pyth. theorem)
1 ) ____
2 ) ( PS
R
We have
60°
P
60°
60°
S
2
2
1
sin 30 PS
PS
1
cos 60
RP
2
RS
3
cos 30
PR
2
RS
tan 60 3
PS
1
PS
tan 30 RS
3
RP
2
2
2
3
sin 60 RS
Q
PR
2
3
or
3
The table below summarizes the trigonometric
ratios of the special angles 30°, 45° and 60°.
Trigonometric
ratio
θ
30
45
60
sin θ
1
2
1
2
or
2
2
3
2
cos θ
3
2
1
2
or
2
2
1
2
tan θ
1
3
or
3
3
1
They are useful when we need to find
the values of trigonometric expressions
involving special angles.
3
Without using a calculator, find the value of the expression sin
30° tan 60° + sin 60°.
1
33
sin 30 tan 60 sin 60 33
2
22
3
3
2
2
3
Follow-up question 15
Find the values of the following expressions without using a
calculator.
sin 45cos 45
1
2
(a)
(b) tan 30
cos 60
tan 2 60
Solution
1
1
sin 45cos 45
2
2
(a)
1
cos 60
2
1
2
1
2
1
Follow-up question 15 (cont’d)
Find the values of the following expressions without using a
calculator.
sin 45cos 45
1
2
(a)
(b) tan 30
cos 60
tan 2 60
Solution
(b)
2
1
1
2
tan 30
2
tan 60 3
1 1
3 3
2
3
1
3
2
Since
exact
valuesangles
of the in
we
canthe
find
the acute
trigonometric
ratios ofequations
special
simple
trigonometric
angles are
known,
without
using
a calculator.
For example:
(a) 2cos 1
1
cos
2
60
(b)
2 tan 2 sin 45
2 tan 2
1
cos 60° = __
2
2
2
2 tan 2
tan 1
45
tan 45° = 1
Follow-up question 16
Find the acute angles in each of the following equations
without using a calculator.
1
(a)
2sin cos 45
(b)
tan sin 60 0
2
Solution
(a)
2sin cos 45
2
2
1
sin
2
2 sin
30
Follow-up question 16 (cont’d)
Find the acute angles in each of the following equations
without using a calculator.
1
(a)
2sin cos 45
(b)
tan sin 60 0
2
Solution
1
(b)
tan sin 60 0
2
1
3
tan
0
2
2
1
3
tan
2
2
tan 3
60
Example 25
Find the values of the following expressions without using a calculator.
(a)
cos 60 tan 30 tan 60
(b)
tan 45 4 sin 30
cos 2 30
(c)
tan 60 sin 60 sin2 45
1 4
(b)
Solution
(a)
1 1
3
2
3
1
1
2
(c)
1
2
cos 60 tan 30 tan 60
1
2
tan 45 4 sin 30
2
cos 2 30
3
2
1 2
3
4
4
3 1
tan 60 sin 60 sin 2 45 3
2 2
3 1
2 2
1
2
Example 26
Find the acute angle in each of the following equations without
using a calculator.
(a)
(b)
cos cos2 45
3 tan ( 10) 1
Solution
(a)
cos cos 2 45
1
cos
2
1
cos
2
60
2
(b)
3 tan ( 10) 1
1
tan ( 10)
3
10 30
40
Example 27
Referring to the figure, find the lengths
of the following line segments without
using a calculator. (Leave your answers
in surd form.)
(a)
AC
(b)
DC
Solution
(a)
Consider △ABC.
AC
sin 45
BC
AC BC sin 45
2
10
cm
2
5 2 cm
(b)
Consider △ACD.
AC
DC
AC
DC
tan 60
5 2
cm
3
tan 60
Finding Trigonometric Ratios by
Constructing Right-Angled Triangles
4
If sin = 5, how can I find cos and
tan ? Do I need to evaluate first?
You can find cos and tan
by the following steps without
evaluating .
Step 1
Construct a right-angled triangle ABC with
A =θand B = 90°.
Step 2
5
opposite side of θ
4
Since sinθ= , we set BC = 4 and AC = 5.
5
5
hypotenuse
Step 3
Find the unknown side AB by Pythagoras’ theorem.
AB AC 2 BC 2
52 42
3
4
3
Step 4
Find the other two trigonometric ratios
by their definitions.
AB
cos
AC
3
5
BC
tan
AB
4
3
In general, if one of the trigonometric
ratios of an acute angle θ is given, we
can follow these steps to find the other
two trigonometric ratios without
evaluating θ.
5
4
3
Follow-up question 17
It is given that tanθ= 0.5, whereθis an acute angle. Find the
values of sinθand cosθwithout evaluatingθ. (Give your answers
in surd form.)
Solution
5
1
tan 0.5
10 2
1
Construct △ABC as shown with tanθ= .
2
By Pythagoras’ theorem,
BC AC AB
2
12 2 2
5
2
C
5
1
A
2
B
Follow-up question 17 (cont’d)
It is given that tanθ= 0.5, whereθis an acute angle. Find the
values of sinθand cosθwithout evaluatingθ. (Give your answers
in surd form.)
Solution
AC
By definition, sin
BC
1
5
or
5
5
2
5
5
1
AB
cos
BC
C
A
2 5
or
5
2
B
Example 28
2
, where is an acute angle. Find the values of
5
cos and tan without evaluating . (Leave your answers in surd
form.)
It is given that sin
Solution
2
Construct △ABC as shown with sin .
5
By Pythagoras’ theorem,
AB
AC 2 BC 2
52 2 2
21
By definition,
AB
cos
AC
21
5
tan
BC
AB
2
21
2 21
or
21
Example 29
It is given that cos 0.25, where is an acute angle. Find the values
of sin and tan without evaluating . (Leave your answers in surd
form.)
Solution
cos 0.25
25
100
1
4
1
Construct △ABC as shown with cos .
4
By Pythagoras’ theorem,
BC
AC 2 AB2
42 12
15
By definition,
BC
sin
AC
15
4
BC
AB
15
1
15
tan
Example 30
40
It is given that tan
, where is an acute angle. Find the value of
9
sin + cos without evaluating . (Give your answer in fraction.)
Solution
40
Construct △ABC as shown with tan
.
9
By Pythagoras’ theorem,
AC
AB 2 BC 2
92 402
1681
41
By definition,
sin
cos
∴
BC
AC
40
41
AB
AC
9
41
40 9
sin cos
41 41
49
41
Trigonometric Identities
Basic Trigonometric Identities
Complete the
table. What
can you find?
sin
cos
θ
sin θ
cos θ
tan θ
30°
1
2
3
2
1
3
1
3
1
45°
1
2
1
2
1
1
1
60°
3
2
1
2
3
3
1
sin 2 cos 2
I find that
2
3045
cos 2 30 1
I find that sinsin
60
30
45
tan
30
2
2 60
sincos
4560
cos
45
1
45
30
sin 2 60 cos 2 60 1
sin
Is
tan always true?
cos
Is sin 2 cos 2 1 always true?
Let’s study the following proof.
Consider the right-angled triangle ABC as shown.
b
a
b
sin , cos , tan
c
c
a
Then
(i)
b
sin
c
a
cos
c
b
a
tan
Consider the right-angled triangle ABC as shown.
b
a
b
sin , cos , tan
c
c
a
Then
2
(ii)
b
a
sin 2 cos 2
c c
2
b2 a2
c2
c2
2
c
1
c2 = b2 a2 (Pyth. theorem)
We have the following two basic
trigonometric identities.
sin θ
tan θ
cos θ
sin2θ cos 2θ 1
Note that
sin2θ cos 2θ 1 can also be written as
sin2θ 1 cos 2θ
or
cos 2θ 1 sin2θ.
Simplify the following expressions.
(b) 1 tan 2
(a) tan
sin
(a) tan tan 1
sin
sin
sin
1
cos sin
1
cos
sin
tan = _____
cos
Simplify the following expressions.
(a) tan
(b) 1 tan 2
sin
2
sin
2
(b) 1 tan 1
cos 2
sin
tan2 = (tan )2 and tan = _____
cos
cos 2 sin 2
cos 2
12
cos
cos2 sin2 = 1
Follow-up question 18
Simplify the following expressions.
(a)
(1 sin 2 ) tan 2
(b)
cos 2 1
1 sin 2
(b)
cos 2 1
1 sin 2
Solution
(a)
(1 sin 2 ) tan 2
cos 2 tan 2
2
sin
2
cos
cos 2
sin2
(1 cos 2 )
1 sin 2
sin 2
cos 2
tan 2
Example 31
Simplify the following expressions.
(a)
sin
tan cos 2
(b) 4 sin 2 4 cos 2
(c)
6 3 sin 2
4 cos 2 4
Solution
(a)
(b)
sin
sin
2
tan cos sin cos 2
cos
sin
sin cos
1
cos
4 sin 2 4 cos 2 4(sin 2 cos 2 )
4(1)
4
(c)
6 3 sin 2 6 3(1 cos 2 )
2
2
sin
cos
1)
(∵
2
2
4 cos 4
4 cos 4
3 3 cos 2
4 cos 2 4
3(cos 2 1)
4(cos 2 1)
3
4
Example 32
Simplify the following expressions.
sin 2
(a) 1
tan 2
sin cos
(b)
1 sin 2
Solution
(a)
sin 2
1
2
1
1 sin
2
tan
sin 2
cos 2
2
cos
2
1 sin
sin 2
1 cos 2
sin 2
(b)
(∵ sin 2 cos 2 1 )
sin cos sin cos
2
1 sin
cos 2
sin
cos
tan
Example 33
8
It is given that sin
, where is an acute angle.
17
(a)
By using the trigonometric identities, find the values of cos
and tan .
(b)
Hence, find the value of
5 sin 2 cos
.
tan
(Give your answers in fractions.)
Solution
(a)
∵
cos 2 1 sin 2
∴
cos 1 sin 2
8
1
17
1
64
289
225
289
15
17
2
sin
tan
cos
8
17
15
17
8
15
(b)
5 sin 2 cos
tan
8 15
5 2
17 17
8
15
40 30
17 17
8
15
10
17
8
15
75
68
Trigonometric Ratios of Complementary
Angles
Consider the right-angled triangle ABC as shown.
a
sin θ =
sin
b
c
cos θ =
cos
b
a
tan θ =
tan
c
c
sin
sin (90°
(90° θ ) =
b
a
cos
cos (90°
(90° θ ) =
b
c
tan
tan (90°
(90° θ ) =
a
=
=
=
1
Using the trigonometric identities, find the acute angle in each
of the following.
1
(a) sin θ cos 35
(b) tan ( 90 θ )
tan 27
(a)
sin cos 35
sin ( 90 35)
sin 55
55
cos = sin (90° )
Using the trigonometric identities, find the acute angle in each
of the following.
1
(a) sin θ cos 35
(b) tan ( 90 θ )
tan 27
1
(b) tan ( 90 θ )
tan 27
1
1
1
tan (90° ) = _____
tan
tan tan 27
tan tan 27
27
Using the trigonometric identities, find the acute angle in each
of the following.
1
(a) sin θ cos 35
(b) tan ( 90 θ )
tan 27
(b) Alternative Solution
tan ( 90 θ )
1
tan 27
tan (90 θ ) tan (90 27)
90 θ 90 27
27
1
tan (90° ) = _____
tan
Follow-up question 19
Using the trigonometric identities, find the acute angle in each
of the following.
(a) cos (90 ) sin 42
(b) tan tan 35 tan 45
Solution
(a) cos (90 ) sin 42
(b) tan tan 35 tan 45
sin sin 42
tan tan 35 1
1
tan
tan 35
42
tan (90 35)
tan 55
55
Example 34
24
It is given that tan
, where is an acute angle. By using the
7
trigonometric identities, find the values of sin and cos . (Give your
answers in fractions.)
Solution
∵
∴
24
tan
7
sin 24
cos
7
7 sin 24 cos (*)
49 sin 2 576 cos 2
49(1 cos 2 ) 576 cos 2
625 cos 2 49
49
2
cos
625
7
cos
25
(∵ sin 2 cos 2 1 )
∵
∴
7 sin 24 cos
7
7 sin 24
25
24
sin
25
(from (*) )
Example 35
It is given that cos
1
, where is an acute angle. Using the
3
trigonometric identities, find the value of 3 cos2 sin2 .
(Give your answer in fraction.)
Solution
3 cos 2 sin 2 3 cos 2 (1 cos 2 )
4 cos 2 1
2
1
4 1
3
4
1
9
5
9
(∵ sin 2 cos 2 1 )
Proofs of Simple Trigonometric Identities
We have
learnt
fivetotrigonometric
can use
them
prove other
identities.
trigonometric identities.
sin
cos
(ii) sin2 cos 2 1
(i) tan
(iii) sin(90 ) cos
(iv) cos( 90 ) sin
(v) tan( 90 )
1
tan
Prove that
1
cos
sin
.
sin
tan
1
L.H.S.
sin
sin
R.H.S.
1 sin 2
sin
cos 2
sin
cos
cos
tan
∵ L.H.S. R.H.S.
1 sin2 = cos2
cos
sin
1
cos
tan
cos
tan
1
cos
______
_____
= sin
sin
_____
cos
1
= _____
tan
1
cos
sin
sin
tan
Follow-up question 20
1
Prove that 1 2
tan 2 θ.
sin (90 θ )
Solution
L.H.S. 1
1
1
sin 2 ( 90 θ )
R.H.S. tan 2
1
cos 2
L.H.S. R.H.S.
(1 cos 2 )
cos 2
sin 2
cos 2
tan 2
1
1
2
tan
2
sin ( 90 )
Example 36
Find the acute angle in each of the following equations by using the
trigonometric identities.
(a)
1
tan
tan 66
(b) cos ( 30) sin
2
Solution
1
tan
tan 66
tan tan (90 66)
(a)
tan tan 24
∴
24
(b)
cos ( 30) sin
2
cos ( 30) cos 90
2
∴
30 90
3
60
2
40
2
Example 37
Simplify the following expressions.
(a) sin (90 ) cos (90 ) tan (90 )
(b) cos (90 ) sin sin 2 (90 )
Solution
(a)
(b)
1
sin (90 ) cos (90 ) tan (90 ) cos sin
tan
1
cos sin
sin
cos
cos
cos sin
sin
cos 2
cos (90 ) sin sin 2 (90 ) sin sin cos 2
sin 2 cos 2
1
Example 38
Find the values of the following expressions.
1 sin 2 37
(a)
sin 2 53
(b) cos 21 tan 69 sin 21
Solution
(a)
1 sin 2 37 cos 2 37
2
sin 53
sin 2 53
sin 2 (90 37)
sin 2 53
sin 2 53
sin 2 53
1
1
sin 21
(b) cos 21 tan 69 sin 21 cos 21
tan (90 69)
1
cos 21
sin 21
tan 21
1
cos 21
sin 21
sin 21
cos 21
cos 21
cos 21
sin 21
sin 21
cos 21 cos 21
0
Example 39
Prove the following trigonometric identities.
sin cos
1
(a)
cos sin sin sin (90 )
(b) sin tan
1
cos
cos
Solution
(a)
sin cos
L.H.S.
cos sin
sin 2 cos 2
sin cos
1
sin cos
R.H.S.
1
sin sin (90 )
1
sin cos
∵
∴
L.H.S. R.H.S.
sin cos
1
cos sin sin sin (90 )
(b)
1
cos
cos
1 cos 2
cos
sin 2
cos
sin
sin
cos
sin tan
R.H.S.
L.H.S. sin tan
∵
L.H.S. R.H.S.
∴
sin tan
1
cos
cos
Example 40
Prove that (sin cos )2 1 2 sin cos .
Solution
L.H.S. (sin cos ) 2
sin 2 2 sin cos cos 2
(sin 2 cos 2 ) 2 sin cos
1 2 sin cos
R.H.S. 1 2 sin cos
∵ L.H.S. R.H.S.
∴ (sin cos ) 2 1 2 sin cos
Extra Teaching Examples
Example 7 (Extra)
In △ABC, ∠ADC = 90, AB = 7.5 cm,
AC = 6.5 cm and AD = 5.5 cm. Find ∠BAC,
∠B and ∠C correct to 1 decimal place.
Solution
∵
AD
sin B
AB
5.5 cm
7.5 cm
∴
B 47.2 (cor. to 1 d.p.)
∵
AD
sin C
AC
5.5 cm
6.5 cm
∴
C 57.8 (cor. to 1 d.p.)
In △ABC,
( sum of △)
BAC 180 B C
180 47.2 57.8
75.0 (cor. to 1 d.p.)
Example 13 (Extra)
In △ABD, ∠D 90, AB 11 cm, AC
7.8 cm and AD 6 cm. Find ∠BAC correct
to the nearest 0.1.
Solution
∵
∴
AD
cos DAC
AC
6 cm
7.8 cm
DAC 39.7
∵
AD
cos DAB
AB
6 cm
11 cm
∴
DAB 56.9
∴
BAC DAB DAC
56.9 39.7
17.2 (cor. to the nearest 0.1)
Example 19 (Extra)
In △ABC, ∠ABD = ∠DBC = 25, ∠C = 90
and BC = 3.5 m. Find AD correct to 2 decimal
places.
Solution
In △ABC,
∵
∴
AC
tan ABC
BC
AC
tan (25 25)
3.5 m
AC 3.5 tan 50 m
In △BCD,
∵
∴
∴
DC
tan DBC
BC
DC
tan 25
3.5 m
DC 3.5 tan 25 m
AD AC DC
(3.5 tan 50 3.5 tan 25) m
2.54 m (cor. to 2 d.p.)
Example 21 (Extra)
The figure shows a quadrilateral ABCD
with ∠A = 115, ∠B = 90, ∠C = 80,
AB =13 cm and AD = 18 cm. Find BC
correct to the nearest cm.
Solution
Draw DF BC and AE DF as shown in
the figure.
In △ADE,
cos DAE
AE
AD
∴
AE AD cos DAE
18 cos (115 90) cm
18 cos 25 cm
∴
BF AE 18 cos 25 cm ……(1)
DE
sin DAE
AD
∴ DE AD sin DAE
18 sin (115 90) cm
18 sin 25 cm
DF DE EF
DE AB
(18 sin 25 13) cm
In △CDF,
DF
tan 80
CF
∴
DF
CF
tan 80
18 sin 25 13
cm
tan 80
……(2)
∴
BC BF CF
18 sin 25 13
18 cos 25
cm
tan 80
20 cm (cor. to the nearest cm)
Example 22 (Extra)
In the figure, sector OPQ is inscribed in
rectangle ABCD. Given that AB =10 cm and
BC = 14 cm, find the area of sector OPQ
correct to 1 decimal place.
Solution
In △OBQ,
OB
cos QOB
OQ
10
cm
2
14 cm
∴ QOB 69.1
QOB POA
POQ QOB POA 180 (adj. s on st. line)
POQ 2QOB 180
POQ 2(69.1) 180
POQ 41.8
∴
41.8
14 2 cm 2
Area of sector OPQ
360
71.6 cm 2 (cor. to 1 d.p.)
Example 27 (Extra)
The figure shows two shadows
AD and BD of a tree CD at
9:00 a.m. and 4:30 p.m.
respectively. If the height of the
tree is 8 m, find the distance
between A and B. (Leave your
answer in surd form.)
Solution
Consider △ACD.
CD
tan 45
AD
CD
AD
tan 45
8m
1
8m
Consider △BCD.
CD
tan 30
BD
CD
BD
tan 30
8m
1
3
8 3 m
Distance between A and B AD DB
(8 8 3 ) m
8(1 3 ) m
Example 38 (Extra)
Find the value of 3 3 tan 2 26 sin 2 64 3 cos 2 26 .
Solution
3 3 tan 2 26 sin 2 64 3 cos 2 26
3 3 tan 2 26 cos 2 (90 64) 3 cos 2 26
sin 2 26
2
2
3 3
cos
26
3
cos
26
2
cos 26
3 3 sin 2 26 3 cos 2 26
3 3(sin 2 26 cos 2 26)
3 3(1)
0
Example 39 (Extra)
Prove that 1 cos 2 (90 ) sin 2 tan 2 (90 ) .
Solution
L.H.S. 1 cos 2 (90 )
1 sin 2
cos 2
∵
∴
R.H.S. sin 2 tan 2 (90 )
1
sin 2
tan 2
1
2
sin
sin 2
cos 2
2
cos
2
sin
sin 2
cos 2
L.H.S. R.H.S.
1 cos 2 (90 ) sin 2 tan 2 (90 )