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CHAPTER 15: TRIGONOMETRY
Important Concepts: Trigonometrical Ratios
Exercise 1: Solution :In Diagram 1, ABC is right-angled triangle.
Trigonometry II 1
Adjacent side
Hypotenuse
B
A
C
Sin = =
Cos = =
Tan = =
Opposite side
17 cm15 cm
8 cm
Sin = =
Cos = =
Tan = =
Example 1 :The figure shows a right-angled triangle.
A
BC
12 cmx
5 cmDiagram 1
13 cm
x
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Determine the value of sin x, Cos x and Tan x.
Exercise 2 : Solution :
In Diagram below, ABD is a straight line
Trigonometry II 2
Solution :
AB = 3 x 4 = 12 cm
AC = 5 x 4 = 20 cm
DC = 12 cm
EC = 13 cm (Pythagoras Theorem)
Therefore, AE = AC EC = 20 13
= 7 cm
Example2 :
In the diagram, AEC is a straight line. Sin ACB = and tan DCE =The length, in cm, of AE is
12 cm
5 cm
A B
C
D
E
x y
C
A D B 8 cm15 cm
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Given tan x = 1, find the length of CD .
Exercise 3 :
In Diagram below, ABC and BDE are two right-angled triangles.
Solution :
Trigonometry II 3
x
6 cmSolution :
QR = 3 x 2 = 6 cm
PQ = 4 x 2 = 8 cm PR = 10 cm (Pythagoras Theorem) Therefore, sin x =
=
Example 3 :In the diagram below, PQRS is formed from two a right-angled triangles.Given that tan y = , find the value of sin x.
P Q
S
R
y
16 cm
9 cm 8 cm A B D
C
10 cm
E
x
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If AB = 9 cm, BD = 8 cm, DE = 10 cm, and BE = CE, calculate the value of sin x.Example 4
In the diagram above,QRS is a straight line . Given
cos13
5=QPS and tan
4
3= RTS , find
the length of RT in
Solution
Given13
5=QPS , so PS = 13,
Given also4
3= RTS ,then RS = 3, RT = 4, and QS = 6 cm
Using Pythagoras theorem, QS is equal to 12 cm.
So, in ratio 43
tan = RTS equal to126
.Meaning RS = 6 cm and RT = 12 cm .
Exercise4
Based on the figure above, calculate the length of FL , in cm .
Trigonometry II 4
Q
S
PTR
5 cm
H F
L
48o
10 cm
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Example 5
In the diagram above FGH is a straight line . Given that tan 1 HEF , thus tan= HGE
Solution
Since tan 1 HEF = EF HF
= EF
GF HF +
=7
25 +
= 1 HGE is equivalent to EGF ,
Tan EGF =2
7, since HGE is greater than 90o , tan = HGE
2
7
Exercise 5
In the diagram above, BCD is a straight line and AB is perpendicular to line BCD .Find BAC .
Trigonometry II 5
5 cm
H
F
G
E 7 cm
14 cm8 cm
A
B C D30o
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Example 6
In the diagram above ,QRS is a straight line .
Given that sin x o =13
12, find the value of
cos yo .
Solution
sin xo=13
12=
PRQR ,
By using Pythagoras Teorem QP = 5 and QRP is equivalent to o y
So, cos =QRP 13
12=
PR
QR ,
since o y is greater than 90o ,o
y = 13
12
Exercise 6
In the diagram above ,FGH is a straight line . find the value of cos xo .
Trigonometry II 6
Q RS yo
xo
P
E
F G
G
xo
5 cm4 cm
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Exercise :
1 Diagram 1 shows a rectangle PQRS, where diagonalSQ = 17 cm and PQ = 15 cm.
Calculate the value of cos x.
2 Diagram 2 shows a triangle KLM .
Diagram 2
Calculate the length, in cm, of KL if ML = 60 cm and cos x =31 .
3 In Diagram 3,M is the midpoint of BC in a right-angled triangle ABC.
DIAGRAM 13
Trigonometry II 7
x
M
K L
60 cm
B
M
C A
x
17cm
15cm P Q
RS
x
Diagram 1
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Diagram 3
Given that cos x =257 and AB = 50 cm, calculate the length, in cm, of MC .
4 In Diagram 4,MNLP is a trapezium.
Diagram 4
If NL = 2MN , sin x =1715 , NL = 8 cm andOP = 15 cm, find the perimeter, in cm, of
the trapezium.
5 In Diagram 5, BCDE i s a rectangle and ABM is a triangle.
Trigonometry II 8
P
OM
L N
15 cm
x
8 cm
D E
C B
A
M
x
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Diagram 5
AMD and ABC are straight lines and AB = 6 cm, and ED = 20 cm. Given that cos x =31 ,
find the length, in cm, of AD.
6 In Diagram 6, ABCD is a rectangle.
Diagram 6
If AB = 9 cm and BX =32
AB. If BC = 16 cm, and BY =21
BC , find sin y.
7 In Diagram 7 cos x =135 .
B
x y
A C D
Diagram 7
If ACD i s a straight line and AC = 5 cm, find the length of BD when sin y = .2
1
8 In Diagram 8 ABC is an equilateral triangle.C
X
Trigonometry II 9
B A X
C D
Y
y
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yA B
Diagram 8
If AXY is a right-angled triangle, find the value of sin y.
9 In Diagram 9, ABCD is a trapezium.15 cm
D C
24 cm
yA B
22 cmDiagram 9
If AD = 24 cm, DC = 15 cm and AB = 22 cm, find the value of cos y.
10 In Diagram 10, the area of square ABCD is 49 cm and the area of triangleCDE is 84cm.
A D
x
B C E
Diagram 10Given that BCE is a straight line, find the value of cos x.
Trigonometry II 10
Y
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TRIGONOMETRY II
Trigonometry II 11
1. Theunit circle is the circle with radius 1 unit and its centre at origin.
2.
a)Quadrant Angle
I 0 < < 90II 90 < < 180III 180 < < 270IV 270 < < 360
b) sin = y = y1
cos = x = x1tan = y
x
1
1
-1
(x, y)
y
y
x
x
All +sin +
cos +tan +
3. 0 90 180 270 360
sin 0 1 0 -1 0cos 1 0 -1 0 1tan 0 Undefined 0 Undefined 0
30 45 60sin 1 / 2 1/ 2 3 / 2cos 3 / 2 1/ 2 1 / 2tan 1 / 3 1 3
4.
Quadrant II1800-Quadrant I
Quadrant III
- 1800Quadrant IV3600 -
900
1800
2700
0, 36
y
x3600
y
x3600
y
x36001800 270090027001800900900 1800 2700
y = sin x y = cos x y = tan x
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15.1 Identifying the Quadrants and the Angles in A Unit Circle.
1. The x-axis and the y-axis divides the unit circle with centre origin into 4 quadrants as shown inthe diagram below
y1 90
180 -1 II I 1 0 O 360 XIII IV
-1 270
Examples and exercises :
State the quadrant for the following angles in the table below.Angle Quadrant Angle Quadrant42 I 19 70 265 100 II 289 136 126 197 303 205 80 275 150 354 212
REMEMBER
QUADRANT I 0 < < 90QUADRANT II 90 < < 180
QUADRANT III 180 < < 270QUADRANT IV 270 < < 360
Trigonometry II 12
1
-1
1 1
-1-1
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15.1 a) Determine whether the values of
a) sin b) cos c) tan are positive or negative if
oooooo and 360270,270180,18090
y1 90
180 -1 Sin + ALL 1 0O 360 X
Tan + Cos +
-1 270
Examples :
i) Sin 142 ii) cos 232 iii) tan 299
142 is in quadrant II cos 232 is in quadrant III tan 299 is in quadrant IVSin is positive in Quadrant II Cos is negative in quadrant III tan is negative in quadrant IV
Exercises :
Angle Quadrant Value (Positive/ Negative)Sin Cos Tan
75 I + + +120 II + - -
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160 200 257 280 345
15.1 b)Find the values of the angles in quadrant I which correspond to the following values of angles in other quadrants.
The relationship between the values of sine, cosine and tangent of angles in Quadrant II, III andIV with their respective values of the corresponding angle in Quadrant I is shown in the diagram
below :
QUADRANT II QUADRANT III QUADRANT IV( 90 180 ) ( 180 270 ) (270 360 )
Sin = sin ( 180 - )Cos = cos ( 180 - )Tan = tan ( 180 - )
Sin = - sin ( - 180 )Cos = -cos ( - 180 )Tan = tan ( - 180 )
Sin = - sin ( 360 - )Cos = cos ( 360 - )Tan = - tan ( 360 - )
Example :
120
Sin 120 = sin 60Cos 120 = - cos 60Tan 120 = - tan 60
EXERCISES :
Find the values of the angles in quadrant I which correspond to the following values of angles inother quadrants.
ANGLE CORRESPONDING ANGLE IN QUADRANTI
Sin 125 Sin = sin ( 180 - 125)= sin 55
Cos 143Tan 98
Trigonometry II 14
230 340
Sin 230 = - sin 50Cos 230 = - cos 50Tan 230 = tan 50
Sin 340 = - sin 20Cos 340 = cos 20Tan 340 = - tan 20
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Sin 200 Sin = - sin ( 200 - )= - sin 20
Cos 245 Tan 190 Sin 285 Sin = - sin ( 360 - )
= -sin 55Cos 300 Tan 315
15.1 c)Find the value of Sine, Cosine and Tangent of the angle between 90 and 360
TIPS :If a calculator is used, press either , or
Followed by the value of the angle and then
Example :
a) sin 145 b) cos 220 = c) tan 92.5
Exercises :
Angle ValueSin 46Cos 57Tan 79Sin 139Cos 154Tan 122Sin 200Cos 187Tan 256
Trigonometry II 15
=
Press displaySin Sin1 sin 14 sin 145 sin 145= 0.573 576
Press displaycos cos2 cos 22 cos 220 cos 220= - 0.951 056
Press displaytan tan9 tan 92 tan 92. tan 92.5 tan 92.5= 0.573 576
SIN COS TAN
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Sin 342Cos 278Tan 305
15.1 d ) Find the angle between 0 and 360 when the values of sine, cosine and tangent areGiven
TIPS :If a calculator is used, press either SIN 1 , COS 1 or TAN 1
Followed by the value of the angle and then
Examples :
a) Sin 1 0. 94 b) Cos 1 -0.64 c) Tan 1 0.625
Press display Press Display Press DisplayShift Sin Sin 1 ShiftCos Cos 1 Shift Tan Tan 10 Sin 1 0 (-) - 0 Tan 1 0. Sin 1 0. 0 Cos 1 -0 . Tan 1 0.9 Sin 1 0.9 . Cos 1 -0. 625 Tan 1 0.6254 Sin 1 0.94 64 Cos 1 -0.64 = 32.00= 70.05 15 = 129.79
Sin 1 0. 94 = 70.05 Cos 1 -0.64 = 129.79 Tan 1 0.625 =32.00
Exercises :
VALUE ANGLESin 1 0.7654Sin 1 -0.932
Trigonometry II 16
=
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Sin 1 0.1256Cos 1 0.4356Cos 1 -0.6521Cos 1 -0.7642Tan 1 -1.354Tan 1 0.7421Tan 1 1.4502
15.1 e) Determine The Value Of Sin , Cos And Tan For Special Angles
A A45
3030
2 2 1 23
4560 60
B D C B 1 C1 1
Using The Right-Angled Triangle Bad, Using Isosceles Triangle
Sin 30 =21 Sin 45 =
Cos 30 =2
3 Cos 45 =
Tan 30 =3
1tan 45=
sin 60 =2
3
cos 60 =
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tan 60 =
15.2 Graphs Of Sine, Cosine And Tangent
15.2.a) For each of the following equations, complete the given table and draw its graph basedon
the data in the table.
i) y = sin x
X 0 45 90 135 180 225 270 315 360 Y
ii) y = cos x
X 0 45 90 135 180 225 270 315 360 Y
iii) y = tan x
X 0 45 90 135 180 225 270 315 360 Y
Examples :
Trigonometry II 18
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The diagram shows graphs y = sin x for 0 360 . Find the value of y when yis a) 90
b) 270c) 360
SOLUTION : a) y = 1b) y = -1c) y = 0
15.3 Questions Based on the Examination Format.
1. Which of the following is equal to cos 35 ?
A. cos 145 C. cos 235 B. cos 215 D. cos325
2. Find the value of sin 150 + 2 cos 240 - 3 tan 225
A. -3.5 B. -1.5 C. 1.5 D. 2.5
3. Sin 30 + cos 60 =
A.41 B.
21 C. 1 D. 0
4. Given that sin 45 = cos 45 = 0.7. Find the value of 3 sin 315 - 2 cos 135
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A. -3.5 B. -1.5 C. 1.5 D. 2.5
5. Given that cos = 0.9511 and 0 360, find the value of
A. 18 B. 162 C. 218 D. 300
6. Given that tan = 05774 and 0 360 , find the value of
A. 30 , 210 B. 152 , 210 C.30 , 330 D. 30 , 150
7. Given that sin = -0.7071 and 90 270, find the value of
A. 135 B. 225 C. 45 D. 315
8. Given that Sin x = 0.848 and 90 x 180 , find the value of x
A. 108 B. 122 C. 132 D. 158
9. Given that tan y = -2.246 and 0 360 , find the value of y
A. 66 , 246 B. 114 ,246 C. 114 , 294 D.246 , 294
10.
y(0,1)
( -1,0) (1,0)O X
(0.87,-0.50)( -1,0)
The diagram shows the unit circle. The value of tan is
A. -1.74 B. -0.57 C. -0.50 D. 0.87
11.y1
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-1 1O X
P-1
The diagram shows the unit circle. If P is (-0.7, -0.6), find the value of Sin
A. -67 B. -
76 C. -0.6 D. 0.6
12y
1
-1 1O X
R (0.8, -0.4)
-1
The diagrams shows a unit circle and R (0.8, -0.4). find the value of cos
A. 0.8 B. 0.4 C. 1 D.8.04.0
13. In the diagram, ABC is a straight line. The value of sin x is
BA C
x
15 8
D
A.15
8 B.17
8 C.17
15 D.15
17
14.T
13 cm5 cm
Q SR
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X 7 cm
U
In the diagram, PQRS is a straight line and R is the mid-point of QS. The value of cos x is
A.13
12 B.25
12 C.25
13 D.25
24
15. P15 cm T 6 cm S
Q
R
In the diagram, PQR and QTS are straight lines. Given that sin TRS =5
3, then
sin PQT =
A.158 B.
17
8
C.15
8 D.17
8
16.
Given that PQR is a straight line and tan x = -1, find the length of PR in cm.
A. 6 B. 8 C. 10 D. 1217.
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In the diagram above, PQR is a straight line. Given that cos53
= SQP , find tan x.
A.21 B.
8
5C.
4
3D.
54
18.
In the diagram above, EFGH is a straight line. If sin53
= JGH , the value of tan x
=
A.5
4B.
21 C.
3
1 D.
5
3
19. Diagram below shows a graph of trigonometric function.
The equation of the trigonometric function isA. y = sin x B. y = -sin x C. y = cos x D. y = -cos x
20.
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The value of cos is
A.3
4B.
5
3C.
5
3 D.
5
4
15.4 PAST YEAR SPM QUESTIONS
Nov 2003, Q11
1. I n Diagram 5, GHEK is a straight line. GH = HE.
7 cm 25 cm
Diagram 5Find the value of tanx
A. 12
5C.
12
13
Trigonometry II 24
F
GE
K H
J
13cm
x
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B. 13
12D.
5
12
Nov 2003, Q12
2. Which of the following graphs represents y = sin x ?
Nov 2004, Q 11
3. In Diagram 5, PRS is a straight line
x
Find the value of coxx =
A.24
7 C. 24
7
Trigonometry II 25
Q
P
7 cm
24 cm
R
S
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B.25
24 D. 25
24
Nov 2004, Q 12
4. Diagram 6 shows the graph of y = sin x.
The value of p is
A. 90 C. 270B. 180 D. 360
Nov 2004, Q13
5. In diagram 7, JKL is a straight line.
Diagram 7
It is given that cosx =13
5 and tan y = 2. Calculate the length, in cm, of JKLA. 22 C. 44
B. 29 D. 58
Nov 2005, Q11
6. It is given that cos = 0.7721 and 180 360. Find the value of
A. 219 27 C. 309 27B. 230 33 D. 320 33
Nov 2005, Q12
7. In Diagram 6, QRS is a straight line.
4 cmQ P
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16 cm
12 cm
x
13 cm
H
E
G
F
3 cm
R Diagram 6
S
What is the value of cos ?
A.5
4C.
5
3
B.5
3D.
54
JUNE 2004, Q13
Diagram 6
8. Diagram 6 shows a quadrilateral EFGH.Find the value of x.
A. 33 01 C. 49 28B. 40 33 D. 50 54
JUNE 2004, Q14
9. In Diagram 7, O is the origin of a Cartesian plane.
Trigonometry II 27
P (-3, 4)
r
y
x0
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1
0
Diagram 7
The value of sinr is
A.5
3
C. 5
3
B. 54
D.
4
3
JUNE 2005, Q12
10. Which of the f ollowing graphs represents y = sin 2x for 0 x 180,?
Trigonometry II 28
-1
1
090 18 00
0
y
x
B
1
0
-1
90 18000
y
x
A
2
1
090 1800
0
y
x
C
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y
N M
P Q
x 0
0
JUNE 2005, Q13
11. In Diagram 5, MPQ is a straight line.
Diagram 5
Given cos x =25
24, find the value of tan y.
A.24
7 B.
7
24 C.
24
7 D.7
24
JUNE 2005, Q11
12. Given cos x
= - 0.8910 and 0
x
360
, find the values of x.A 117 and 243 C. 153and 207
B 117 and 297 D 153and 333
NOV 2005, Q11
13. It is given that cos = -0.721 and00 360180
. Find the value of .
A. 19o 27
B. 230o33
C. 309o27
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D. 320o33
NOV 2005, Q12
14. In Diagram 6, QRS is a straight line
Diagram 6
What is the value of cos 0
A.54
B.5
3
C.53
D.5
4
JUNE 2006, Q11
15. Diagram 5 shows a rhombus PQRS
Diagram 5
It is given that QST is a straight line and QS = 10cm.Find the value of tan xo.
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A.135 C.
125
B.12
13D.
5
12
JUNE 2006, Q12
16. Which of the following represents part of the graph of y = tan x?
A. C.
B. D.
JUNE 2006,Q13
17. In Diagram 6, PQR and TSQ are straight lines.
Find the length of ST , in cm.
A. 2.09 C. 3.56
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B. 3.44 D. 4.91
NOV 2006, Q11
18. In Diagram 5, S is the midpoint of straight line QST.
The value of cos xo is
A.34 C.
43
B.5
4D.
5
3
NOV 2006, Q12
19. In Diagram 6, MPQ is a right angled triangle.
It is given that QN = 13cm, MP = 24cm and N is the midpoint of MNP.Find the value of tan y0.
A.13
5 C.
1312
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B.125
D.1213
NOV 2006, Q1320. Which of the following represents the graph of y = cos x for 00 1800 x ?
A.
B.
C.
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D.
JUNE 2007, Q13
21. In Diagram 7, SPQ and PRU are right angle triangles. STQ and PTU are straight lines.
It is given that cos yo =13
12and PQ = QR . Calculate the length incm, of PTU
Trigonometry II 34
S
R
U
P Q
T
20O yo
Diagram 7
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A. 25.54
B. 27.67
C 65.94
D. 70.17
JUNE 2007,Q14
22. In Diagram 8. PRS is a straight line,
If tan xo
= 43
, then the value of h is
A. 5
B. 15
C. 16
D. 20
JUNE 2007 , Q15
23, Which of the following represents the graph of y = sin x for 0o x 369o
A.
Trigonometry II 35
P 12 cm Q
xoR
S
Diagram 8
h cm
0.5
0 90 1800
y
x360O
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B.
C.
D.
NOV 2007, Q11In diagram 6, USR and VQTS are straight lines,
Trigonometry II 36
0.5
0 60O180O
y
x360O
0.5
0 45O180O
y
x360O
1
0
y
x360O
-1
180o
V Digram 6
U
S
R
xo
T
P
Q
yo
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It is given that TS = 29 cm, PQ = 13 cm, QR = 16 cm and sin xo =17
8 ,
Find the value of tan yo
A. 512
B.12
5
C.12
5
D.5
12
NOV 2007, Q12
24.
In Diagram 7, O is the origin and JOK is a straight line on a Cartesian plane.The value of cos is
A. -5
4
B. -5
3.
Trigonometry II 37
J0 x
y
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C.5
3
D.5
4
NOV 2007, Q1325. Which of the following graphs represents y = Sin x for 00 1800 x ?A.
B.
C.
Trigonometry II 38
1
0 90O 180O
y
x
-1
1
0 90O 180O
y
x
-11
0 90O 180O
y
x
-1
1
0 90O 180
y
x
-1
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JUNE 2008,Q 12
27. In Diagram 7, RTU is a right angled triangle RST and TUV are straight lines
It is given that RS = 28 cm, TU = 15 cm and tan RUV = -5
12
Find the length, in cm, of SU.
A. 23
B. 22.63C. 17
D 15.73
JUNE 2008,Q13
28. Given that sin x = -2
1, 27090 0 x find the value of 3 cos x.
A. -2
3
B.2
3
C 23
D. 23
JUNE 2008,Q14
28. Which graph represents y = cos x for 00 3600 x ?
Trigonometry II 39
T U V
R
S
Digram 7
1
0180o
y
x360O
-1
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A.
B
C
D
NOV 2008,Q11
30. Digram 6 shows a right angled triangle PQR.
Trigonometry II 40
1
0 180o
y
x360O
-1
1
0 180o
y
x360O
-1
1
0180
o
y
x360O
-1
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Given sin xo =21 , find the value of h.
A. ok 30tan
B. k tan 30O
C. ok 60cos
D. k cos 60o
NOV 2008, Q13
31. Which of the following represents the graph of y = tan x for 00 3600 x ?
A.
B.
Trigonometry II 41
P k Qxo
R
h
Diagram 6
1
0
y
x360O
-1
180o
1
0
y
x360O
-1
180o
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C
D.
ANSWERS
Chapter 15 Trigonometry
Exercise 1:
Sin x =1312 , Cos x =
135 , Tan x =
512
Exercise 2: CD = 17 Exercise 3: Sin x =9
12=
3
4Exercise 4 : FL = 9cm
Exercise 5 : BAC = 28.96
Exercise 6 :5
3
Exercise:
1. Cos x =17
82. KL = 20 cm 3. MC = 12.5 cm
4. Perimeter = 48 5. AD = 18 + 60 = 78 6. Sin y =10
8 or 5
4
7. BD = 24 8. Sin y = Sin 30Sin 30 = 0.5
9. Cos y =96.22
7
10. Cos x =25
24
Trigonometry II 42
1
0
y
x360O
-1
180o
1
0
y
x360O
-1
180o
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15.1a)Angle Quadrant Angle Quadrant42 I 19 I70 I 265 III100 II 289 IV136 II 126 II197 III 303 IV205 III 80 I275 IV 150 II354 IV 212 III
15.1 b.Angle Quadrant Value (Positive/ Negative)
Sin Cos Tan75 I + + +120 II + - -160 II +
200 III +257 III +280 IV + 345 IV +
15.1 cANGLE CORRESPONDING ANGLE IN QUADRANT
ICos 143 Cos 37Tan 98 Tan 82
Cos 245
Cos 65
Tan 190 Tan 10Cos 300 Cos 60Tan 315 Tan 45
15.1 dAngle Value
Sin 46 0.7193398Cos 57 0.5446390Tan 79 5.1445Sin 139 0.6560
Cos 154 -0.8987Tan 122 -1.6003Sin 200 -0.3420Cos 187 -0.9925Tan 256 4.01078Sin 342 -0.30901Cos 278 0.13917Tan 305 -1.42814
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15.1 eVALUE ANGLE
Sin 1 0.7654 49..94Sin 1 -0.932 68.74
Sin1
0.1256 7.215Cos 1 0.4356 64.17Cos 1 -0.6521 49.29Cos 1 -0.7642 40.16Tan 1 -1.354 53.55Tan 1 0.7421 36.57Tan 1 1.4502 55.411
15.3: EXAMINATION FORMAT QUESTIONS No Answer No Answer
1 D 11 C2 A 12 B3 C 13 C4 B 14 D5 A 15 B6 A 16 D7 A 17 A8 B 18 C9 C 19 B10 B 20 D
15.4
Questions Answers Questions Answers NOV 2003, Q11 A NOV 2003, Q12 D
JUN 2004, Q13 C NOV 2004, Q11 DJUN 2004, Q14 B NOV 2004, Q12 A
NOV 2004, Q13 BJUN 2005, Q12 C NOV 2005, Q11 AJUN 2005, Q13 B NOV 2005, Q12 DJUN 2005, Q11 CJUN 2006, Q15 D NOV 2006, Q11 BJUN 2006, Q16 A NOV 2006, Q12 CJUN 2006, Q17 C NOV 2006, Q13 BJUN 2007, Q13 A NOV 2007, Q11 DJUN 2007, Q14 D NOV 2007, Q12 BJUN 2007, Q15 D NOV 2007, Q13 CJUN 2008, Q12 C NOV 2008, Q11 BJUN 2008, Q13 A NOV 2008, Q12 A
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JUN 2008, Q14 C NOV 2008, Q13 D