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Module PMR
Trigonometry 169
Summary of the trigonometry
1. Tangent of an acute angle 2. Sine of an acute angle 3. Cosine of an acute angle 4. The values of Tangent, Sine and Cosine
C A B
a. Sin θ = Hypotenuse
Opposite
b. Cos θ = Hypotenuse
Adjacent
c. Tan θ = Adjacent
Opposite
A. Calculating the value of sine, cosine and tangent of an angle. Example :
1. In the diagram, AC = 26 cm and BC = 24 cm. Find the value of sinθ . C
24 cm 26 cm
B A
Opposite Side Hypotenus
Adjacent Side
θ
CHAPTER 11 :TRIGONOMETRY
θ
Module PMR
Trigonometry 170
Solution:
Sin C = AC
AB
= 26
10
= 13
5
Exercise : Calculate the value of sin θ for each of the following triangles. 1). R θ
6 cm P Q 10 cm
2). C B 5 cm 13 cm A
3). A 15 cm B C 8 cm
4). R 25 cm P Q 24 cm
AB2 = AC2 – BC2
= 262 - 242
= 676 – 576 = 100
AB = 100 = 10 cm
θ θ
θ
θ
Module PMR
Trigonometry 171
Example
1. In the diagram, AB = 15 cm and AC = 9 cm. Find the value of cos θ .
C B 9 cm
15 cm A Solution:
Cos θ = AB
BC
= 15
12
= 5
4
Exercise : Calculate the value of cos θ for each of the following diagram. P 1). R 17cm Q
2). R 12 cm Q 5 cm P
BC2 = AB2 – AC2 = 152 - 92 = 225 – 81 = 144
BC = 144 = 12 cm
8 cm
θ θ
θ
Module PMR
Trigonometry 172
3). K M 15 cm 8 cm L
4). A 12 cm B 15 cm C
Example
1. Find the value of tan θ in the diagram as shown. P
10 cm Q 6 cm R Solution :
Tan θ = PQ
QR
= 8
6
= 4
3
θ
θ
θ
PQ2 = PR2 – QR2
= 102 - 62
= 100 – 36 = 64
PQ = 64 = 8 cm
Module PMR
Trigonometry 173
Exercise : Calculate the value of tan θ for each of the following diagram. 1). P 25 cm 7 cm Q R
2). A B 13 cm 12 cm C
3). A 15 cm B 9 cm C
4). L 5 cm 13 cm M 4 cm N P
θ
θ
θ
θ
Module PMR
Trigonometry 174
B. Calculating the lengths of sides of a triangle from the tangent, sine and cosine of an angle. Example : Find the value of x in the following triangle. B x cm A C 15 cm
tan θ = 5
3
Solution :
tan θ = 5
3
5
3=AC
BC
5
3
15=x
155
3 ×=x
cmx 9=
θ
Module PMR
Trigonometry 175
Exercise. Find the value of x in each of the following triangles. 1). A x cm B 10 cm C
tan 5
2=θ
2). P 6 cm x cm Q R
tan 2
1=θ
3).
tan 24
7=θ
4).
tan 24
7=θ
θ
θ
θ
C
B A
x cm 48 cm
x cm
14cm
C
B A
θ
Module PMR
Trigonometry 176
5). K L x cm 12 cm M
sin 3
1=θ
6). R x cm S 6 cm T
sin 5
3=θ
7).
sin 5
3=θ
8).
sin 5
3=θ
θ θ
θ
x cm
15 cm
C B
A
θ x cm
9 cm C B
A
Module PMR
Trigonometry 177
9). R x cm S 14 cm T
cos 7
5=θ
10). A x cm 16 cm B C
cos 8
5=θ
11).
cos 13
5=θ
12).
cos 13
5=θ
θθ
θ
26 cm
x cm C B
A
θ
x cm
52 cm
C
B A
Module PMR
Trigonometry 178
Common Errors
1. Find the value of cos x in the diagram as shown.
Error Correct step
Cos BC
CD=θ
8
7=
Cos CE
AC=θ
13
12=
2. Find the length of AC if sin 13
5=θ
Error Correct Steps
Sin 13
5=θ
13
5=AC
BC θtan=
AC
BC
13
510 =AC
Sin 13
5=θ
13
5=AB
BC
13
510 =AB
7 cm
6 cm
E
D
C 8 cm B 4 cm A
θ
θ
10 cm
B
C A
Module PMR
Trigonometry 179
5
1310×=AC
cmAC 26=
5
1310×=AB
cmAB 26= 222 1026 −=AC 576= 576=AC cmAC 24=
Extra Exercise 1. In diagram below, PQR is a straight line and T is the midpoint of straight line QTS. (a) Given that tan x 1=� , calculate the length of QTS. (b) State the value of cos y � .
P Q S
S
T
4 cm 15 cm x y
Module PMR
Trigonometry 180
2. In diagram below, S is the midpoint of straight line TSQ.
(a) Given that cos y5
4=� , calculate the length of TQ.
(b) Find sin x � . 3. In diagram below, C is the midpoint of straight line BD.
Given that sin x13
5=� , find the length of DC.
T S
R
Q
P
y
x
8 cm
16 cm
x
A
B C D
5 cm
Module PMR
Trigonometry 181
4. Diagram below shows two right angled triangles, PQR and RST. PRS and QRT are straight line.
Given that cos x17
15=� and sin y5
4=� .
(a) Find the value of tan x. (b) Calculate the length, in cm, of QRT.
P
R T
S
Q
x
y
17 cm
10 cm
Module PMR
Trigonometry 182
5. Diagram below shows two right angled triangles, ABC and ACD.
It is given that cos x5
4=� and tan y3
2=� .
(a) Find the value of sin x. (b) Calculate the length, in cm, of AD.
A B
C
D
x
y
8 cm
Module PMR
Trigonometry 183
6. Diagram below shows two right angled triangles, PQR and RST. PRS and QRT are straight lines.
It is given that sin x5
4=� and cos y13
12=� .
(a) Find the value of tan x. (b) Calculate the length, in cm, of QRT. 7. Diagram below show two right angled triangles, ABD and BCD.
Given that tan y5
12=� , find the value of cos x.
T
y
S R P
x
Q
13 cm
10 cm
D
A
B
C x
y
15 cm
13 cm
Module PMR
Trigonometry 184
8. In diagram below, JKL and KMN are straight line. M is the midpoint of the line KN.
Given that cos y5
4=� , find the value of sin x.
9. The diagram shows two right-angled triangles PQR and RSQ.
Given that sin x 13
5= ,
a). find the value of tan y, b). calculate the length of PR, in cm.
5 cm 8 cm
N
J
M
K L
y x
x
y 13 cm
16 cm Q P
R S
Module PMR
Trigonometry 185
10. In the diagram, tan x 5
12= .
Find the length of RS, in cm
x
12 cm 15 cm
T
S R Q
Module PMR
Trigonometry 186
PMR past year questions 2004 1). In Diagram 1, C is the midpoint of the straight line BD.
Find the value of tan xo
2005 2). Diagram 1 shows a right angled triangle EFG and DEF is a straight line. G 4 cm Find the value of cos x°
5 cm 13 cm
B C D
x
x°
2 cm D E F
Module PMR
Trigonometry 187
2006 3). Diagram 3 shows two right angled triangles, DAB and CDB.
It is given that tan y12
5=� and sin x2
1=� .
(a) Find the value of cos yo
(b) Calculate the length, in cm, of BC.
5 cm
B
C
y
x
Module PMR
Trigonometry 188
2007 4). Diagram 10 shows twi right angled triangles, PQT and SQR. PQR and TQS are straight lines. T
It is given that sin x13
5=� and cos y5
3=� .
(a) Find the value of tan x° (b) Calculate the length, in cm, of PQR.
15 cm
5 cm
y x P
Q R
S
Module PMR
Trigonometry 189
2008 5). Diagram 20 shows a right angled triangle PQR.
It is given that cos y13
5=� , calculate the length, in cm, of PQ.
y
P
Q R
39 cm
Module PMR
Trigonometry 190
CHAPTER 11 : TRIGONOMETRY ANSWERS
SIN θ
1). sin 5
4=θ
2). sin 13
12=θ
3). sin 17
8=θ
4). sin 25
7=θ
COS θ
1). cos 17
15=θ
2). cos 13
12=θ
3). cos 17
15=θ
4). cos 5
3=θ
TAN θ
1). tan 24
7=θ
2). tan 5
12=θ
3). tan 3
4=θ
4). tan 12
5=θ
VALUE OF x 1). x = 4 cm
2). x = 3 cm
3). x = 14 cm
4). x = 4.083 cm
5). x = 4 cm
6). x = 8 cm
7). x = 9 cm
8). x = 12 cm
9). x = 10 cm
10). x = 10 cm
11). x = 10 cm
12). x = 20 cm
EXTRA EXERCISE 1). (a) QTS = 8 cm
(b) Cos y17
15=�
2). (a) TQ = 20 cm
(b) Sin x5
3=�
3). DC = 6 cm
4). (a) Sin x17
8=�
(b) QRT = 14 cm
Module PMR
Trigonometry 191
5). (a) sin x5
3=�
(b) AD = 15 cm
6). (a) Tan x3
4=�
(b) QRT = 18 cm
7). Cos x5
3=�
8). Sin x13
12=�
9). (a) tan y 4
3=
(b) PR = 20 cm
10). RS = 4 cm
PMR past year questions 2004 2005
1). Tan x5
6=� 2). Cos x2
1−=�
2006 2007
3). a). Cos y13
12=� 4). a). tan x o = 12
5
b). BC = 26 cm b). PQR = 21 cm 2008 5). PQ = 15 cm