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Trigonometry / Class X . sem.2 / page - 1
CHAPTER 6
6.1 TRIGONOMETRY RATIOS
Consider the figure of a right triangle below :
Defined that :

r
Sine  =
y
r
or
x
or
r
y
Tangen  =
or
x
x
Sine  =
or
r
y
Cosine  =
or
r
Cosine  =
y

x
y
r
x
Cos  =
r
y
Tan  =
x
x
Sin  =
r
y
Cos  =
r
Sin  =
For instance, in the following right triangle :
5
3
5
4
Cos  =
5
4
5
3
Cos  =
5

Sin  =
4

Tan  =
Sin  =
4
3
Tan  =
3
4
Sin,Cos and Tan are trigonometric ratios, while the other trigonometric ratio the other
trigonometric ratio are represented as follows :
1
1
r
then Cosec  = y =
Sin 
y
r
1
1
r
or Sec  =
then Sec  = x =
x
Cos 
r
1
x
1
or Cot  =
then Cot  = y =
Tan 
y
x
r
y
r
Secant 
=
x
x
Cotangents  =
y
Cosecant  =
Cot 
x
=
or Cosec  =
r
y
r

cos 
sin 
and
Tan  =
y
x
r
r

sin 
cos 
Trigonometry / Class X . sem.2 / page - 2
So that in the last triangle we have :
Cosec  =
5
4
Cosec  =
5
3
,
5
3
Sec  =
, Cot  =
3
4
Cot  =
4
3
And also
,
Sec  =
5
4
,
EXERCISE 1
I.In each right triangle below , find :
(a) Sin 
(d) Sin 
(b) Cos 
(e) Cos 
(c) Tan 
(f) Tan 
(1)
(g) Cosec 
(h) Sec 
(i) Cot 
(2)
20
(3)

21

( j ) Cosec 
( k ) Sec 
(l) Cot 


17

60

8
II.On the ∆ PQR , RQ altitude ( RQ  PS )
Find the trigonometry ratio as follows :
R
(a) Sin P
(b) Sec P
(d) Sin R1
(e) Tan R1
(g) Cosec S (h) Cos S
(j) Sec R2
(k) Tan R2
12
P
Q
(c) Cot P
(f) Cos R1
(i) Cot S
(l) Sin R2
S
6.2 DETERMINING OTHER TRIGONOMETRIC RATIOS WHEN
ONE RATIO IS GIVEN
Although the trigonometric ratio can be found for obtuse angles ( > 90o ) here we still
consider that the angles are acute and are in a right triangle.
Principally , if any trigonometric ratio is given then we can determine the corresponding
right triangle , and from that triangle we can find the other trigonometric ratios.
Trigonometry / Class X . sem.2 / page - 3
EXAMPLES
1.If sin x =
1
, then find : (a)cos x
3
(b)tan x
(c)sec x
(d)cot x
Answer :
A right triangle of sin x =
1
is represented beside.
3
3
1
x
8
(a)cos x =
2
2
3
2.If tan x =
2
6 then find : (a) cos x
7
(b)tan x =
1
2
4
(c)sec x =
(b)sin x
3
2
4
(d)cot x = 22
(c)cot x
Answer :
We may write : tan x =
73
x
7
26
2
6 , and the corresponding right triangle is shown beside
7
7
2
73 (b)sin x =
438
(a) cos x =
73
73
(c) cot x =
7
6
12
Trigonometry / Class X . sem.2 / page - 4
EXERCICE 2
Complete the table for the acute angle x follows.
No
1
Sin x
2
Cos x
……..
Tan x
………
Cot x
……….
Cosec x
……..
Sec x
……..
…….
………
……..
……..
………
……….
………..
3
2
……
3
4
3
……
…..
5
6
4
……..
……..
………
……….
3
……..
5
6
……..
……..
……..
………
…….
……..
………
2,5
……..
…….
2
……..
7
…….
……..
………
……..
…...
2
3
5
7
6.3 USING TRIGONOMETRIC RATIO TO FIND LENGHT OF SIDE
TRIANGLES
EXAMPLES
B
In the triangle beside given that sin  BAC = 0,6 and
7 cm .Find the length.
1.
7 cm
A
C
Answer: Since sin  BAC =
BC
, then BC = AB , sin  BAC = 7. 0,6 = 4,2 cm
AB
Trigonometry / Class X . sem.2 / page - 5
2.
B
Given that tan  BAC =
1
and BC =
3
20
Find the length of AB.
A
C
Answer :
10
A
1
3
The right triangle of tan =
C
we find sin =
1
is shown beside , where
3
1
10
BC
BC
then AB =
AB
sin BAC
Since sin  BAC =
=
20 cm
1
10
=
200 cm
= 102 cm
3.In the figure beside the triangle ABC is right angled at C
AC = 6 cm , BC = 8 cm and CD is perpendicular to AB
Find the length of AD.
B
D
A 6 cm
Answer :
Cos  BAC =
8 cm
C
6
3
AC

=
10 5
AB
Cos  CAD = Cos  BAC =
Cos  CAD =
3
5
3
AD
so that AD = AC , cos  CAD = 6 .
= 3,6 cm
5
AC
Trigonometry / Class X . sem.2 / page - 6
4.
B
In the triangle beside given that :
12 cm
Cos  BAC =
15 cm
9
3
, cos  ACB =
, AB = 12cm
16
4
BC = 15 cm.
Find the length of AC.
A
C
AD
, then AD = AB
AB
27
9
Cos  BAD = 12 .
=
cm
4
16
Since cos  BAD =
Answer :
B
12 cm
A
15 cm
D
C
Cos BCD = cos  ACB =
3
4
DC
45
3
, then DC = BC , cos BCD = 15 .
=
cm
BC
4
4
27
45
So that : AC = AD + DC =
+
= 18 cm
4
4
Since cos BCD =
EXERCISE 3
1.The figure shows a string is stretched between peak of a banner pole and the
land where tan  =
2
5 and length of the string is 18 meters , then find the height of
5
the banner pole .

Trigonometry / Class X . sem.2 / page - 7
2. In the triangle given that cot  =
1
7
3
find : (a) altitude (b) area of triangle , by the base of side 16 cm
12 cm

16 cm
3.In the triangle DEA right angled at A given that DA = 30 cm and EA = 16 cm .
The lines AA1 and A2A3 are perpendicular to DE, while the line A1A2 is perpendicular
to DA.
E
A1
A3
D
A
A2
Find the lengths of : (a) AA1
(b) A1A2
(c) A2A3
(d) DA2
(e) AA2
(f) DA3
(g) DA1
(h) A1A3
4.In a cone of base radius 3 cm , given that sin  =
Find : (a) height
3 CM
3 CM
(b) volume
1
2.
10
(c)surface area of the cone.
Trigonometry / Class X . sem.2 / page - 8
5.
C
D
E
B
ABC triangle is right angled at A . If BC = p ,
AD is perpendicular at BC , DE is perpendicular
at AC ,  B = 
Prove that : DE = p sin2. Cos 
A
6.PQR triangle is right angled at R and sin P . cos Q =
tan P
3
. Find the value
.
5
tan Q
7.Given that ABC triangle with the side of BC = 3 cm , side AC = 4 cm and sin A =
8.
1
2
C

A

D
B
Given that ABC triangle is right angled at B . The point D is located at AB so that AD = 1
. If  BAC =  ,  BDC = . Prove that : BC =
sin  . sin 
sin   sin 
6.4 The SPECIAL ANGLES
B
The triangle ABC beside is a equilateral triangle
Of sides AB = BC = AC = 2 cm
By recalling the properties of an equilateral triangle , then
BAC = ABC = ACB = 60o
if BM is perpendicular to AC , then we find that ABM =
CBM = 30o and AM = MC = 1 cm
30o 30o
60o
A
60o
M
By Pythagorean: BM =
C
AB 2  AM 2

2 2  12  3 cm
Trigonometry / Class X . sem.2 / page - 9
1
BM
3
, then we obtain sin 60o =
2
AB
AM
1
Since cos  BAM =
, then we obtain cos 60o =
AB
2
BM
Since tan  BAM =
, then we obtain tan 60o= 3
AM
1
AM
3
Since cot  BAM =
, then we obtain cot 60o=
3
BM
AB
Since sec  BAM =
, then we obtain sec 60o = 2
AM
Since sin  BAM =
2
AB
3
, then we obtain cosec 60o =
3
BM
AM
1
Since sin  ABM =
, then we obtain sin 30o =
AB
2
1
BM
3
Since cos  ABM =
, then we obtain cos 30o =
2
AB
1
AM
3
Since tan  ABM =
, then we obtain tan 30o =
3
BM
Since cosec  BAM =
So we have know sin ,cos and tan of angles 30o and 60o
Now consider the isosceles triangle DEF right angled at E , where DE = EF = 1 cm
F
450
2
1 cm
450
E
1 cm
D
We have  DEF =  EDF = 45o , and Pythagorean : DF =
1
EF
2
then sin 45o =
2
DF
1
ED
2
Since cos  DEF =
then cos 45o =
2
DF
EF
Since tan  DEF =
then tan 45o = 1
ED
ED
Since cot  DEF =
then cot 45o = 1
EF
DF
Since sec  DEF =
then sec 45o = 2
ED
Since sin  DEF =
DE 2  EF 2  12  12
Trigonometry / Class X . sem.2 / page - 10
Since cosec  DEF =
DF
then cosec 45o = 2
EF
The below is tabulated the values of trigonometric ratios we have already calculate.
Sin
0
0o
30o
Cos
1
1
2
45o
1
60o
2
1
1
2
2
1
Tan
0
1
3
Cot

2
3
3
3
2
3
Cosec

2
3
1
1
3
1
2
2
2
3
2
3
2
2
3
2
90o
Sec
1
1

0
0
3
3

1
EXERCISE 4
1.The triangle DEF is equilateral , DF = 12 cm as shown
Find : (a) altitude (b) area of the triangle DEF
E
D
F
2. Given that  CBD = 600 ,  DAC = 300 and AD = 10 cm. Find the length BC and AC
D
10 cm
A
B
2.Find the value :
a.
cot 45 0
cos ec 90 0
b.sin 600.cos 300 + cos 600 . sin 300
c.cos2300 + sin2600 + cos2450
C
Trigonometry / Class X . sem.2 / page - 11
d.sec245 + cot2600 + cosec2300
e.
sin 2 60 0. tan 2 30 0  2. sec 60 0. cos 0 0
cos ec 2 45 0. cot 60 0
3.Calculated the unknown element in the triangle below .
a.
b.
10
300
x=?
600
600
x
2
c.
300
d.
300
4
x=?
5
450
x
0
0
30
45
4.Given ∆ ABC with base AC = 20 cm . If  A = 450 and lengght of AB = 10 cm ,
determine the area for ∆ ABC.
5.
D
6 cm
Given that ABCD rectangle with  A =  D =
900 and AD = CD = 6 cm. Determine the
Length of BC and AB sides.
C
6 cm
300
A
B
6.A trapezoid ABCD with a base AB , the length of side BC = 7,5 cm , the height of the
trapezoid is 6 cm ,  A = 500 ,  B = 0 , the length of AD = p cm , determine the
valued of p and  .
Trigonometry / Class X . sem.2 / page - 12
7.
The triangle wxy is right angled at w and  wyx = 60o
Meanwhile ww1 and w1w3 are perpendicular to xy
w1w2 is perpendicular to wy
If xy = 1 dm then express in cm units the length of :
(a) ww1
(b) w1w2
(c) w2w3
(d) ww2
(e) w2y
(f) yw3
(g) w1w2
x
w1
w3
600
w
w2
y
8.If in the figure AB =
3
2 cm and CD = 2 cm , then find the length of :
2
D
(a)BE
(d)CH
(b)EF
(e)BF
(c)EH
B
F
E
450
0
30
A
H
C
How to remember the rule of sign ?
II
I
Sin
Cosec
all
III
Tan
Cot
IV
Cos
Sec
If we consider the ratio sin , cos , tan cot ,
Sec and cosec , then :
1) In the first quadrant ( 0o – 900) : all
of ratios are positive
2)In the second quadrant ( 90o-180o) :
only sin and cosec is positive
3)In the third quadrant (180o-270o) :
only tan and cot is positive
4)In the fourth quadrant (270o-360o) :
Only cos and sec
Trigonometry / Class X . sem.2 / page - 13
6.5.Trigonometric Ratios Formulas of
Related Angles
1.Ratios of ( 180 -  )0 ( the second quadrant b)
sin(180 - )0=
Y
P’(-a,b)
P(a,b)
( 180 -  )0

X
a
csc(180 - )
0
sec (180 - )
= sin 
r
a
= - cos 
cos(180 -  )0= 
r
b
tan(180 - )0 = 
= - tan 
a
cot(180 - )0 =
b
r
=
= cosec 
= - cot 
b
0
r
= - sec 
a
=
10
2.Ratio of ( 180 +  )0 ( the third quadrant )
Y
P (a,b)
( 180 +  )0
sin ( 180+ )0 = 
cos ( 180+ )0 =  a = - cos 

tan ( 180+ )0 =
r
b
a
cot ( 180+ )0 =
a
= cot 
b
X
P’(-a,-b)
csc ( 180+ )0 =
sec ( 180+  )0 =
b
= - sin 
r


r
b
r
a
= - cosec 
= tan 
= - sec 
11
Trigonometry / Class X . sem.2 / page - 14
3.Ratio of ( 360 -  )0 ( the fourth quadrant )
Y
sin ( 360 -  )0 = 
b
= - sin 
r
cos ( 360 - )0 =
a
= cos 
r
P(a,b)
(360 -  )0

tan ( 360 - )0 =  b = - tan 
a
a
P’(a , -b )
cot ( 360 -  )0 =  = - cot 
r
b
csc ( 360 -  )0 =
= - csc 
b
X
r
a
sec ( 360 - )0 =
= sec 
12
4.Ratio ( - 0 ) ( the fourth quadrant )
y
P (a,b)
Sin (- 0 ) = 
x

-
P ‘(a,-b)
Cos (- 0 ) =
a
r
Tan (- 0 ) = 
b
= - sin 
r
= cos 
b
= - tan 
a
a
= - cot 
b
r
Sec( - 0 ) =
= sec 
a
r
= - csc 
Csc( - 0 ) = 
b
Cot (- 0 ) = 
13
Trigonometry / Class X . sem.2 / page - 15
5.Ratio of ( 90 -  )0 ( the first quadrant )
a
= cos 
r
b
cos ( 90 - )0 =
= sin 
r
a
tan ( 90 - )0 =
= cot 
b
b
cot ( 90 -)0 =
= tan 
a
sin ( 90 -  )0 =
y
P ‘ ( b,a)
r
( 90 -  )0

P ( a,b)
r
x
csc ( 90 -  )0 =
r
= sec 
a
sec ( 90 -  )0 =
r
= csc 
b
14
6.RATIO ( 90 -  )0 ( the second quadrant )
P ‘(-b,a)
y
a
= Cos 
r
b
=- Sin 
cos ( 90 +  )0 = 
r
a
tan ( 90 +  )0 = 
= - cot 
b
sin ( 90 +  )0 =
( 900 +  )
P ( a,b)

x
b
a
= - tan 
r
b
a
csc ( 90 +  )0 =
r
= - csc 
cot ( 90 +  )0 = 
sec ( 90 +  )0 = 
= sec 
15
Trigonometry / Class X . sem.2 / page - 16
7.Ratio ( 270 -  )0 ( the third quadrant )
y
a
Sin (270 - )0 = 
= - cos 
r
P (a,b)
(2700 -  )
x

Cos (270 - )0 = 
Tan (270 - )0 =
Cot (270 - )0 =
P ‘(-b,-a)
Sec (270 - )0 = 
r
b
Cec (270 - )0 = 
b
r
a
b
= - sin 
= cot 
b
= tan 
a
= - csc 
r
a
= - sec 
16
8.Ratio ( 270 +  )0 ( the fourth quadrant )
P (a,b)
(2700 +
)

P ‘(b,-a)
a
r
b
Cos (270 + )0 =
r
a
Tan (270 + )0 = 
b
b
Cot (270 +  )0 = 
a
Sin (270 + )0 = 
= - cos 
= sin 
= - cot 
= - tan 
r
= csc 
b
r
Csc (270 + )0 = 
= - sec 
a
Sec ( 270 +  )0 =
17
Trigonometry / Class X . sem.2 / page - 17
If an angle is greater than 360o
Here , we should subtract the angle by
360o or its multiples , so that the angle
lies between 0o - 360o
Then :
(a) sin  = sin ( + k.360o ) , k  B
(b) cos  = cos (  + k.360o ) , k  B
(c) tan  = tan (  + k.360o ) , k  B
(d) cot  = cot ( + k.3600 ) , k  B
(e) sec  = sec ( + k.3600 ) , k  B
(f) csc  = csc ( + k.3600 ) , k  B
The below is tabulated the values of trigonometric ratios we have already
calculate.
o
0
30o
Sin
0
Cos
1
1
2
45o
1
60o
2
1
1
2
2
1
Tan
0
3
1
3
1200
1350
1500
1800
2100
2250
2400
2700
3000
3150
3300
3600
Sec
1
2
3
2
3
Cosec

2
3
1
1
3
1
2
2
2
3
2
3
2
2
3
2
90o
Cot

3
1
0

0
3

1
3
Trigonometry / Class X . sem.2 / page - 18
EXERCISE 5
1.Simplify :
(a)
sin (180 0   )
sin ( 90 0   )
(b)
cos ( 90 0   )
sin (90 0   )
(c)
tan (90 0   )
sec(90 0   )
(d)
sec(90 0   )
cos ec(180 0   )
(e)
sec( 270 0   )
cot(360 0   )
(f)
sin( 360 0   )
sin( 270 0   )
2.Find the value:
(a) sin 240o.tan(-1500 ) + sec (-2100 ). cot (- 330o)
(b)
cos 480 0 . cot 135 0  ( 32 ) sin (1050 0 ). sec 600 0
 2 . tan (840 0 ). cos ec(1200 0 )
3.If tan 40o = a , ( a  R and a  0) express the following of a
tan 140 0  tan 130 0
(a)
1  tan 140 0. tan 130 0
tan 220 0  tan 130 0
(b)
tan 230 0  tan 320 0
4.Solve the following for 0o < x < 360o
(a) cos x = - 0,35
(b) sin x = - 0,65
(c) tan x = 081
5.Solve the following equations for 0o < x < 360o
(a) 5 sin2x = 2 sin x
(b) 9 tan x = cot x
(c) 6 sin2x + 7 sin x +2 = 0
(d) sin (x + 60o) = -0,75
(e)cos2x = 0,6
(f) 5 sin2x = 2
6.Find the solutions set :
(a) 2 sin x – 1 = 0 for -5400 < x < 7200
(b) cot x + 1 = 0
for -1080 < x < 540
(c) 3sec x + 23 = 0 for -5400 < x < 5400.
7. Find out the solution set of :
: 2 sin2(270 – x )o = 1 + cos ( 180 + x )o for – 360o  x  360o
8.Prove that at any ABC triangle applies sin ( A + B ) = sin C
9.Point out that :
Cos 200 + cos 400 + cos 600 + .... + cos 1600 + cos 180 0 = - 1
10.Point out that :
tan 150.tan 160.tan 170....... tan 730.tan 740.tan 750 = 1
Trigonometry / Class X . sem.2 / page - 19
11. Given that cot x =
1
3

)+
and   x 
3

.Find the value
2
2 Sin ( x -  ) +
cos ( x +
3
Cosec (  - x )
2
2
12.If cos A = (a) tan A
(b) sec A
13.If tan x = (a) sin x
14.If sin  =
4
,  A quadrant II then find :
5
(c) cosec A
4
( 90o < x < 180o ) .Find :
3
(b) cot
(c) sec x
24
12
dan cos  =
( ,  is acute ) Find :
25
13
(a) sin  . cos  - cos  . sin 
(b) cos  . cos  - cos  . sin 
(c)
tan   tan 
1  tan  . tan 
15.Given sin 650 = 0,901 ; cos 650 = 0,755 ; tan 650 = 0,870 , calculate the value of
sin 1150 , cos 1150 and tan 1150
16.Given sin 350 = 0,574 ; cos 350 = 0,819 and tan 350 = 0,700 , calculate :
Sin 1450 + cos 2150 – tan 3250
17.Show that :
cot 99 0
cos 378 0
a.
x
cos 810
cos 198 0
=1
18.Find the value x : cot ( 400 – 2x ) = tan ( 3x – 50 )0 for 00  x  900
6.6. TRIGONOMETRIC IDENTITIES
B
1
The triangle ABC beside is right angled at C ,
 BAC = x and AB = 1 so that AC = cos x
and BC = sin x
Since AC2 + BC2 = AB2 then :
cos2x + sin2x = 1 …….(I)
A
C
sin x
BC
Since tan  BAC =
, then tan x =
………………..(II)
AC
cos x
Dividing equation (I) by cos2x we obtain :
Trigonometry / Class X . sem.2 / page - 20
1 + tan2x =
1
cos 2 x
or 1 + tan2x = sec2x ………..(III)
while dividing equation ( I ) by sin2x , we obtain :
cot2x + 1 =
1
sin 2 x
or cot2x + 1 = cosec2x
EXAMPLES
Example 1. Show that : (cos x + 2 sin x)2 + (2cos x – sin x)2 = 5
Answer :
(cos x + 2 sin x)2 + (2cos x – sin x)2 =
= cos2x + 4 cos x.sin x + 4 sin2x + 4 cos2x – 4 cos x.sin x + sin2x
= cos2x + sin2x + 4 (cos2x + sin2x ) + 4 (cos x.sin x - cos x.sin x )
= ( 1 )
+4( 1 )
+ 4.0
= 5
Example 2 . If cos x =
1
, then find (a) sin x
4
(b) tan x
(x is acute )
Answer :
(a) Since cos2x + sin2x = 1 , then sin x =
(b) tan x =
sin x
=
cos x
1
4
15
1
4
1  cos 2 x  1  161 
= 15
EXERCISE 6
1..Prove the following identities:
(a) (a cos x + b sin x )2 + (b cos x – a sin x )2 = a2 + b2
(b) tan x + cot x = cosec x . sec x
(c)
cos x
sin x
1  tan 2 x


cos x  sin x cos x  sin x
1  tan 2 x
(d) 1 
cos ec 2 x  sec 2 x
 2 sin 2 x
2
2
cos ec x  sec x
(e) ( tan x + cot x )2 – (tan x - cot x )2 = 4
(f) sin4x + cos4x = 1 – 2 sin2x.cos2x
(g)
1
1

 2 sec 2 x
1  sin 
1  sin 
1
15
4
Trigonometry / Class X . sem.2 / page - 21
(h) tan2 = sin2 ( 1 + tan2 ).
2.If sin  + cos  =
1
(  acute ) , calculate
3
a. tan  + cot 
b. sin3 + cos3
c. sin4 + cos4
3. Prove the following identities:
cos x
= - sec x
sin 2 x  1
tan 4 x  1
b.
= tan2x – 1
2
sec x
cos  

c.  tan  
 sin  = tan 
1  sin  

a.
d.cos2 ( 1 – tan4 ) + sin2 = tan2
sin 3   cos 3 
= sin  - cos 
1  sin  . cos 
cos 2   cos 2  . sin 2 
f.
= cot 4 
sin 2   cos 2  . sin 2 
e.
6.7 RADIAN MEASURE
DEFINITIOS OF RADIAN :
1 cm
2 cm
1 cm
2 cm
3 cm
3 cm
Trigonometry / Class X . sem.2 / page - 22
Shows three circles of radius 1 cm , 2 cm and 3 cm having arc lengths of 1 cm , 2 cm
and 3 cm respectively.
r
One radian is the angle
r 1 radian
made by an arc of length
equal to the radius
r
In general , if the arc length is  times the radius , the angle subtended is  radian ,
i.e. angle subtended at centre =
i.e
arc lenght
radius
=
s
r
Example :
Find the angle subtended by an arc the centre :
(a)
(b)
15 cm
1,5 cm
p
2 cm
q
3 cm
Answer :
(a) p =
1.5
 0,75 rad
2
(b) q =
15
 5 rad
2
RELATIONSHIP BETWEEN AND DEGREES
When arc length AB = r ,  AOB = 1 rad
When arc length AB = 2r ,  AOB = 2 rad
When arc length AB = 2r ,  AOB = 2 rad
As the circumference of circle of radius r has length 2r and subtends an angle of
360o
at the centre , therefore :
2 rad = 360o
 rad = 180o
B
O
A
1 rad =
180 0

 57,3o
Trigonometry / Class X . sem.2 / page - 23
1 rad =
1o =
180 0


 57,3o
 0,001745 rad
180 0
The value cannot be found exactly as  is an irrational number.
To convert radian measure to degree measure , we multiply the radian measure
by a factor of
180 0

.
Conversely , to convert degree measure to radian measure, we multiply the degree
Measure by a factor of

.
180 0
Example :
1.Convert the following radians to degree measure :
5
rad
(a)
(b) 0, 8 rad
6
5
5 180 0
rad =
x
 150 0
6
6

180 0
(b) 0 ,8 rad = 0,8 x
= 45,8o
Answer: (a)

2.Find the value of : (a) sin 0,4
(b)cos 2.2
Answer : (a) sin 0.4rad = 0,389
(c) tan
5

4
(b) cos 2.2rad = - 0,589
(c) tan
5
 =1
4
EXERCISE 7
1.Find the angle subtended by an arc at the centre :
(a)
(b)
(c)
9,6 cm
p 4 cm
5 cm
q
8 cm
3 cm
r
Trigonometry / Class X . sem.2 / page - 24
2.Convert the following to degree measure :
(a)
2

3
5
(b)1 rad
6
(c) 0.8 rad
3.Find the values of the following :
(a) sin 2
(b) cos 0,25
(d ) 1,25 rad
(d) tan 1.6
4.Find the value :
sec 34 . sin (  7  )  cos (   ) . tan 4 
7
6
4
3
2
2. cot (  ). cos ec 56 
3
5.Find the solution set :
(a) 3 tan x + 3 = 0 for - 2 < x < 2
(b) cosec x - 2 = 0
(c) 2cos ( x -
for -
5
  x  2
2
2
 ) = 3 for - 3 < x < 3
3
6.In each of the following cases , find the arc length , s .
(a)
(b)
2 cm
1,8 rad s
3000
4 cm
s
7. In the figure a sector OABCB of a circle centre O and radius 5 cm .If the length of the
chord AB is 8 cm , find :
(a)  in radians
(b) the length of the arc ACB
O
5 cm
8. The length of an arc in a circle of radius 5 cm is 6 cm. Find the angle subtended at
centre.
9.The angle subtended at the centre of a circle by an arc of length 2 cm is 60o.
Find the radian of the circle.
10.In the figure the chord AC of a circle centre O and radius 10 cm is 14 cm .
Find :
(a)the angle subtended at the centre of the circle
by the chord.
(b)the length of the arc ABC.
Trigonometry / Class X . sem.2 / page - 25
6.8 TRIGONOMETRIC FUUCTIONS AND THEIR GRAPHS
The function such as f(x) = sin x , f(x) = cos x and f(x) = tan x are called trigonometric
functions and the graphs are represented below .
Properties :
(a) max = 1 , min = -1
(c) period = 360o
(b) sin(-x) = - sin x
y = sin x
1
- 900
1800
00
- 180
900
4500
2700
3600
-1
Properties :
(a) max = 1 , min = -1
(c) period = 360o
(b) cos(-x) = - cos x
x
y = cos x
- 1800
3600
-
900
00
900
1800
2700
x
4500
Trigonometry / Class X . sem.2 / page - 26
Properties :
(a) there are not max and min
(b) tan (-x) = - tan x
(c) line x =  90o ,  270o ,  k.90o
k odd number are asymptotes
(d)period = 180o
Example :
00
Sketch the graph of : (a) y = sin x + 3
(b) y = 2 cos x
(c) y = sin 2x
(d) y = cos ( x – 45o )
450
900
Asymptot
Answer:
(a) It means that all of y in the graph y = sin x is added by 3,so that we obtain the
following graph
y
y = sin x + 3
4
3
- 1800 - 900 00
Properties:
a. max = 1+3 =4
min = -1+3= -2
b. Period = 3600 (constant)
900
1800
2700
3600
x
It is equal to the graph y = sin x is lifted up for 3 units.
(b) It means that all of y in the graph y = cos x is multiplied by 2,so that we obtain the
following graph.
y
2
- 1800
- 900
-2
Properties:
a. max = 2.1 = 2
min = 2.(-1) = -2
b. Period = 3600 (constant)
y = 2 cos x
00
900
1800
2700
3600
x
Trigonometry / Class X . sem.2 / page - 27
(c) Here, each x in the graph y = sin x is divided by 2,as follows
y
Properties:
a. max = 1, min = -1
1
0
b. Period = 360 = 1800
2
- 900
- 450
00
450
900
1350
1800
x
y = sin 2x
-1
In general ,the graph of y = sin kx can be obtained by dividing each x in the graph
y = sin x by k, so that the period be becomes one k-th its former period.
And so is in the graphs of y = cos kx and y = tan kx.
(d) Here you may imagine that each x in the graph y = cos x is added by 450.
y
Properties:
a. max = 1, min = -1
1
b. Period = 3600 (constant)
c.intersects y axis at the
point (00, 1 2 )
2
x
- 450
00
450
900
1350
2250
y = cos ( x – 45 )0
-1
Since when x = 00 then y = cos (00 – 450) = cos 450 = 1 2
2
EXERCISE 8
I.Construct the following graphs for -1800 x 3600 and write down the properties.
1. y = cos x + 2
9. y = cos 1 x
2
2. y = sin x – 2
3. y = tan x – 1
4. y = - 2 cos x
5. y = - sin x
6. y = - tan x
7. y = sin 3x
10. y = sin (x – 450)
11. y = cos (x + 450)
12. y = - 2 tan (x + 450) + 1
13. y = 3 cos x + 2
14. y = 4 sin 2x – 1
15. y = 2 cos ( 2x – 900) + 1
8. y = tan 2x
16. y = 5 – 3 sin ( 2x – 300)
II.Find the maximum , minimum and periods :
1. y = 3 – 2 sin ( 3x – 45 )0
2. y = 3 cos ( 4x – 120 )0 - 5
17. y = 3 cos ( 1 x + 450) – 2
2
18. y = 3 tan ( 3x – 1350) – 1
19. y = sin |x|
20. y = |sin x|
21. y = cos |x|
22. y = |cos x|

23. y = sin x  
2

Trigonometry / Class X . sem.2 / page - 28
6.9 TRIGONOMETRIC EQUATIONS
Examples 1.
Find the values of x that satisfy the equation 2 sin x – 1 =0, if 00< x <3600
Answer
2 sin x = 1
sin x = 1
2
By recalling the graph of y = sin x, then : either x = 300 or 1500. (see the graph below)
Example 2.
Find solution set of : 2 cos2 x – 3 cos x +1 =0, -900  x 3600
Answer
By factoring : (2 cos x – 1)(cos x – 1) =0
First possibility : 2 cos x – 1 =0 or cos x = 1 , gives x = {-600,600,3000}
2
Second possibility : cos x – 1 =0 or cos x =1, gives x = {00,3600}
Union, gives : HP = {-600,00,600,3000,3600}
Example 3.
Find solution set of : sin2 x0 + sin x0 – 2 =0, 00  x  3600
Answer
Factoring: (sin x0 + 2)(sin x0 – 1) =0
First possibility : sin x0 + 2 =0 or sin x0 = -2, is not possible, since minimum value of sin is
only -1.
Second possibility : sin x0 – 1 =0 or sin x0 = 1, gives x = 900
HP : {900}
Example 4.
Find solution set of : cos x – 1 =0, 00 < x < 3600
Answer
HP =  (Empty set)
Why?
You might think that cos x = 1 is satisfied by either x = 00 or x = 3600.
But, both x = 00 and x = 3600 are not included in 00 < x < 3600, as requested by prob.
Trigonometry / Class X . sem.2 / page - 29
6.10 General Solution
The following is the general formula for the solution of trigonometric equation :
sin x = sin 
so x1 =  + k.3600 or x2 = ( 180 -  )0 + k . 3600
x1 =  + k .2 or x2 = (  -  ) + k . 2
cos x = cos  so x1 =  + k. 360o or x2 = -  + k. 360o or x2 = (360 -  )0 +k.3600
x1 =  + k. 2 or x2 = -  + k. 2
or x2 = (2 -  )o + k . 2
tan x = tan  so x =  + k. 180o or x =  + k. with k = 0 ,  1 ,  2 ,  3 , ….
Example :
Determine the solution of the following equation :
a. sin ( 2x – 30 )o - 1
2
3 = 0 for – 3600  x  360o
for - 2  x  2
b. cos 2x = cos x
c. tan ( 2x +  ) = 1
for -   x  
6
for -   x  
d. cosec 2x = sec x
2
2
Answer :
a.sin ( 2x – 30 )o - 1
2
sin ( 2x – 30)0 = 1
2
3 = 0 for – 3600  x  360o
3
2x – 300 = 600 + k.3600
2x = 900 + k. 3600
x = 450 + k.1800
k = - 2  x = - 3150
k = - 1  x = - 1350
k = 0  x = 450
k = 1  x = 2250
or
2x – 300 = 1200 + k .3600
2x = 1500 + k . 3600
x = 750 + k.1800
k = - 2  x = - 2850
k = - 1  x = - 1050
k = 0  x = 750
k = 1  x = 2550
The solution set = { - 3150, - 2850 , - 1350 , - 1050 , 450 , 750 , 2250 , 2550 }
b. cos 2x = cos x for - 2  x  2
2x = x + k. 2 or
2x = - x + k. 2
x = k.2
3x = k. 2
k1 = - 1  x = - 2
x = k. 2
k2 = 0  x = 0
k1 = - 3  x = - 2
k3 = 1  x = 2
k2 = - 2  x = - 4 
3
3
k3= - 1  x = - 2
3
k4 = 0  x = 0
k5 = 1  x = 2
3
k6 = 2  x = 4 
3
Trigonometry / Class X . sem.2 / page - 30
k7 = 3  x = 2
The solution set = { - 2 , - 4  , - 2 , 0 , 2 , 4  , 2  }
3
3
c.tan ( 2x +  ) = 1
3
3
for -   x  
6
2x +  =  + k . 
6
4
2x = 2 + k .
24
x=  +k. 
2
k1 = - 2  x = - 23 
24
k2 = - 1  x = - 11 
24
The sulotion set = { - 23  , 24
24
k4 = 1  x = 13 
24
11  ,  , 13  }
24
24 24
d.cosec 2x = sec x
for -   x  
24
k3 = 0 
2
cosec 2x = cosec (  – x )0
2
x= 
2
cosec 2x = cosec (  - (  - x ) )
2
2x = ( -  + x ) + k.2
2x =  – x + k .2
2
2
3x =  + k.2
x =  + k. 2
2
2
k=0x= 
x =  + k. 2
6
k1 = - 1  x = - 
k2 = 0  x = 
3
2
2
6
The solution set : { -  ,  ,  }
2
6
2
EXERCISE 9
Find the solution sets :
1.2 cos 2x + 3cos x + 1 = 0
, -360o < x < 360o
2.cos2x – sin2x = 0
,-<x<
3.cos (3x – 20)o = - ½ 2
4.tan ( 2x -45 )o + 1 = 0
,
-½<x<
, - < x < 
Trigonometry / Class X . sem.2 / page - 31
, - 540o < x < 360o
5.sin x = tan x
6.tan ( 2x +


) = tan
6
3
, - < x < 
3
  x  2
2
7.sin 2x = cos x
,-
8.cos ( x + 20 )o = cos 70o
, -270o < x < 540o
9.sin (3x – 10 )o = sin 2x
10.sec ( 4x – 45 )o =
11. tan 3x = cot 2x
, - 180o < x < 360o
2
3 , - 180o < x < 180o
3
, -  x 
2
2
, -x
12. tan x = - tan 400
13. cos ( 2x – 30 )0 = sin ( x + 60 )0 ,
- 1800  x  1800
14. ( sin (2x ) + 0,5 ) ( cos x – 1 ) = 0 , - 2  x  2
15.( tan (3x) – 1 ) ( cosec ( 2x -  ) – 2 ) = 0 , -   x  
6.11. SINE FORMULA
C
E
a
It has been mentioned before that sine rule is
use to find out the said length or angle
magnitude of any triangles.
The sine rule used withtout drawing its altitude
as an aid line.
b
A
B
D
c
On ABC triangle , CD represent one of its altitudes , AB = c , BC = a and AC = b
Consider ∆ ADC :
Sin A = CD  Cd = b sin A
b
…..(1)
Trigonometry / Class X . sem.2 / page - 32
Consider ∆ BCD :
Sin B = CD
a
 CD = a sin B …..(2)
a = b
sin A
sin B
b
With the same method for AE altitude will be obtained :
= c
sin B
sin C
a
b
b
c
From
=
and
=
, so the following formula is obtained :
sin A
sin B
sin B
sin C
(1) = (2) so b sin A = a sin B , so that is obtained :
a
b
c


sin A sin B SinC
And the same result is obtained if the triangle ABC is obtuse
Example
In the triangle ABC beside :
 BAC = 45o ,  ACB = 120o ,BC = 6 cm
Find the length of AB.
B
6
450
1200
A
C
Answer:
Sine formula :
AB
BC

0
sin 120
sin 45 0
AB = sin 120o.
=
1
6
3.
1
2
2
2
= 3 cm
Trigonometry / Class X . sem.2 / page - 33
EXERCISE 10
E
6
1.(a) Find the length of EF
(b) Find radius of circumscribed of triangle DEF
2.Find the length of BC
600
D
450
F
C
2 cm
300
450
A
B
3.(a) Find the length of AD and DB
(b) Find radius of circumscribed circle
of triangle ADB
D
300
300
A
B
23 cm
4.On the triangle BDE cos  EBD = 4 , cos  BDE = 3 and ED = 8 cm .
5
5
Find the length of BE.
5. On the triangle ABF ,  AFB = 450 , tan  FAB = 12 , AB = 12 2 cm . Find the
5
length of FB.
6.On the triangle ABC , if a = 16 ,  A = 490 ,  B = 570 . Find the length of b.
7. On the triangle ABC , if a = 25,5 ,  A = 1020 ,  C = 570 . Find the length of c.
8. On the triangle ABC , if b = 16 , c = 23 ,  B = 430 . Find value  C .
9. On the triangle ABC , if a + b + 40 ,  C = 680 and  A = 750 . Calculate  B , a , b
and c.
10.AOn the triangle ABC ,  BAC =  ,  ABC = 150 , AB = 8 2 cm and cos  = - 1 .
2
Find the length of BC.
11.Two ships , S and R , sail 10 km . The position of S is 1100 from R and T is 1700 from
R . Ship T is 2100 from S . Count the distance of T from R and from S.
12.A man is measuring a land . The land side PQ is 520 m.R is measured from P and Q
, and is found  QPR = 490 and  PQR = 780 . Count point R from P and Q.
Trigonometry / Class X . sem.2 / page - 34
13.A and B are the landmark of an area. B is 16 m on the east of A . C is the third
landmark , it is 1270 from A and 2270 from B . count the distance of C to A and C to
B.
D
14.
190
A
From the picture beside , B is
in the hill foot . From B ,
D can be seen with elevation
Angle 370
370
1500 m
B
C
6.12.COSINE RULE
C
b
a
A
D
B
c
Suppose a triangle ABC is given . From the point C , make line Cd that
perpendicular to line AB , thus forming two right-angled triangles ADC and BDC . In
the triangle ADC , the trigonometry result :
Cos A =
AD
AC

AD = AC . cos A = b.cos A
( DC )2 = ( AC )2 – ( AD )2
= b2 – ( b cos A)2
= b2 – b2 cos2A
The triangle BDC satisfies :
( BC )2 = ( BD )2 + ( DC )2
= ( BA – AD )2 + b2 – b2 cos2A
= ( c – b cos A )2 + b2 – b2 cos2A
= c2 – 2 bc cos A + b2 cos2A + b2 – b2 cos2A
= c2 – 2 bc cos A + b2
a2 = b2 + c2 – 2 b c cos A
Trigonometry / Class X . sem.2 / page - 35
In the same way results :
b2 = a2 + c2 – 2 b c cos B
c2 = a2 + b2 - 2 a b cos C
The cosine rule can be used in calculation of triangle cases , if given at least :
a. The length of two sides and the size of angle between them
b. The length of the three sides
Notice that if we rearange the cosine formula , then we obtain :
b2  c2  a2
Cos  =
2 b.c
2
a2  c2  b2
Cos  =
2 a.c
2
a2  b2  c2
Cos  =
2 a.b
2
EXAMPLE:
1.
B
In the triangle ABC beside given that AB = 3 cm ,
AC = 4 cm ,  BAC = 60o
Find the length of BC.
3 cm
Answer :
Cosine formula :
60o
A
C
BC2 = 32 + 42 – 2.3.4 cos 60o = 13
BC = 13
4 cm
2.
B
In the triangle ABC beside AB = 3 cm , AC = 4 cm,
BC = 13 cm.
13 cm
3 cm
A
4 cm
Find :
(a) the size of angle BAC
(b) the size of angle ACB
(c) the size of angle ABC
C
Answer :
(a) cos  BAC =
 BAC = 60o
3 2  4 2  ( 13 ) 2 1

2.3.4
2
Trigonometry / Class X . sem.2 / page - 36
(b) cos  ACB =
4 2  ( 13 ) 2  32

20

5
13
26
2.4. 13
8 13
5
5
13 ) or  ACB = cos-1 (
13 )
 ACB = arc cos (
26
26
32  ( 13 ) 2  4 2
1
(c) cos  ABC =

13
13
2.3. 13
 ABC = arc cos (
1
13 )
13
 ABC = cos-1(
or
1
13 )
13
EXERCISE 11
1.
E
In the triangle DEF beside given that DF = 6 cm
EF = 5 cm and  DEF = 60o
5 cm
Find :
(a) length DE
(b) the size of angle DEF
(c) the size of angle EDF
60o
D
6 cm
F
2.
R
10 cm
In the triangle QRT beside given that QT
= 8 cm, RT = 10 cm, RTQ = 1200.
Find :
(a) length of QR
(b) cos RQT
(c) sin QRT
(d) tan (RQT + QRT)
1200
Q
8 cm
T
3. In the triangle DEF given that DE = 4 cm, EF = 5 cm, DF = 6 cm. Find :
(a) cos DEF, (b) cos EDF ,(c) cos DFE
and determine whether the triangle is acute or obtuse
4. Is the triangle KMN (KM = 5 cm, MN = 9 cm, NK = 7 cm) acute? Explain
Trigonometry / Class X . sem.2 / page - 37
5.
In the triangle beside given that :
sin  = 3 and sin  = 5

4
8
Find the value of cos 


6.
B
In the triangle ABC beside given that AB =7cm
BC = 8 cm, and AC = 9 cm
7 cm
8 cm
A
D
AC = 9 cm
C
The point D is the midpoint of AC, so that BD
is a median.
(a)Find cos BAC (formula(VII))
(b)Find the length of BD (cosine formula in
Δ ABD
(c)Find cos  ACB
(d) Find the length of BD,by cosine formula in Δ BDC.(Notice that the answer must
equal answer b
(e) Find angle ABD
(f) Find angle DBC
7.
B
In the triangle ABC beside given that AB=7cm,
BC = 8 cm, and AC = 9 cm.
The lines BD and AE are medians of triangle ABC.
(a) Find the length of AE
(b) If AE and BD meet at the point T, then find the
Length of AT.
E
T
A
D
C
8.A plane takes off with course of 0840 as far as 280 kilometers , then files with course of
0240 as far 155 kilometers. What is the distance of the plane counted from base ? Find
out the course of the plane order that the plane returns to the first base.
Trigonometry / Class X . sem.2 / page - 38
6.13 TRIANGLE AREAS
Consider the following figure.
Acute triangle
Obtuse triangle
B
B


c

a
c


A
C
b
a

W
W
A
B
If L is area of the triangle ABC, then for both triangles above :
L = 1 .AC.BW
2
AC = b , sin  =
BW
 BW = c sin 
c
L ∆ ABC = 1 b.c sin 
2
In the same way results :
L ∆ ABC = 1 a.b sin 
L ∆ ABC = 1 a c sin 
2
2
If given a side and two angle sizes , the formula to find the area of the triangle are as
follows :
L=
a 2 . sin B . sin C
;
2 sin A
L=
b 2 . sin A . sin C
;
2 sin B
B

A
c 2 . sin A . sin B
2 sin C
In the triangle beside :
2 2 2
cos α  b  c  a
2 bc
2
Since cos  + sin2  = 1. then

c
L=
a

B
Trigonometry / Class X . sem.2 / page - 39
sin2  = 1 
=
(b 2  c2  a 2 )
(2 bc) 2
(2 bc)2  (b 2  c2  a 2 )2
(2 bc) 2
=
(b
2

 c 2  2bc)  a 2 a 2  (b 2  c 2  2bc)
(2 bc) 2

(b  c) 2  a 2  a 2  (b  c) 2 

 

= 
(2 bc) 2
(b c a)(b  c a)(a  b c)(a  b c)
=
(2 bc) 2
(32)
So that
bc sin  = 2  a  b c   b c a   a  c b   a  b c 


2
2


2
2

since area of the triangle ABC is L = 1 bc sin , and if half circumference of triangle is
2
a

b

c
s, that is s =
, then we obtain that :
2
L=
s (s a) (s b) (s c)
This formula is called Heron’s Formula and it tells how to find area of a triangle when the
three sides are given.
If the diagonal of a quadrilateral are known and the angle between the diagonals is also
known , so to count the area of the quadrilateral is applied the following formula.
D
C

P
A
B
The area of ABCD quadrilateral here under is :
L =
1 AC . BD sin 
2
Trigonometry / Class X . sem.2 / page - 40
Proof :
L
 ABC
=
L
 APB
+
L
 BPC
= 1 PA.PB sin ( 1800 -  ) + 1 PB.PC sin 
2
2
= 1 PA. PB sin  + 1 PB.PC sin 
2
2
= 1 PB.( PA + PC ) sin 
2
= 1 PB . AC sin 
2
L
 ACD
=
L
 APD
+
L
 CPD
= 1 PA.PD sin  + 1 PC.PD sin ( 1800 -  )
2
2
= 1 PA.PD sin  + 1 PC . PD sin 
2
2
= 1 PD ( PA + PC ) sin 
2
= 1 PD.AC sin 
2
L
 ABCD
=
L
 ABC
+
L
 ACD
= 1 PB . AC sin  + 1 PD.AC sin 
2
2
= 1 ( AC ( PB + PD ) sin 
2
= 1 AC.BD sin  ( proven )
2
Example 1
Find area of triangle ABC in the figure beside.
B
Answer
L = 1 6 cm . 14 cm . sin 300
14 cm
2
= 21 cm2
0
30
A
C
6 cm
Example 2
Trigonometry / Class X . sem.2 / page - 41
Given that ABCD quadrilateral with Ac = 5 cm ang BD = 8 cm . Determine the area of
ABCD quadrilateral if the angle between AC and BD diagonals is 500
Answer :
L = 1 .AC.BD sin 500 = 1 . 5 . 8 .0,7660 = 15,32 cm2
2
2
Example 3
Determine the area of ∆ ABC , if known b = 6 ,  B = 450 and  C = 600
Answer :
Using sine rule is obtained :
 A = 1800 – ( B +  C )
= 1800 – ( 450 + 600 )
= 750
Using formula
: L=
b 2 . sin A . sin C
6 2. sin 75 0 . sin 60 0
=
= 21.29
2 sin B
2 sin 45 0
So , the area of the triangle is 21.29 area units.
EXERCISE 12
1.
E
The triangle DEF in the figure beside is an
equilateral triangle of side a and of area L.
Prove that : (a) L = 1 a 2 3 , (b) R = 1 a 3
a
D
4
3
F
2.
The figure beside shows a regular hexagon of side
a
Prove that :
a
Area of hexagon = 3 a 2 3
2
Trigonometry / Class X . sem.2 / page - 42
The figure beside shows Δ DEF (DE = 28 cm,
EF = 21 cm, DF = 35 cm) and its circumscribed
circle. Find area of shaded region.
3.
E
 π  22 
7 

D
F
Note : DE is the diameter
4.
Find area of Δ DEB beside.
B
300
300
D
E
10 cm
5.Calculate the area of each part of ABC triangle in the following , if it is know :
a. a = 3 , b = 4 and  C = 450
b. b = 6 , c = 4 and  A = 1500
c. a = 28 ,  A = 530 and  B = 610
d. c = 5 ,  A = 700 and  C = 500
e. a = 7,9 , b = 4,5 and  A = 1000
f. a = 88 , b = 123 and C = 1140
g. a = 12 , b = 14 and c = 20
h. a = 16 , b = 20 and c = 32
6.The area of ABC triangle is 20,7 cm2 , AB side = 6,4 cm and C side = 8,5 cm . Find out
the measure of A angle ( two possibilities )
7.Given that ABCD quadrilateral with AC = 6 cm and BD = 10 cm . Find out the area of
the quadrilateral mentioned if the angle between AC and BD is 600
8.Two contiguous sides of a parallelogram are 8 cm and 6 cm , the side flanked angle is
300 . Calculate the area of the parallelogram.
9.ABCD shape is a trapezoid with AB  DC , AB = 5 cm , BC = 3 2 cm ,  BCD = 450 ,
and  CDA = 600, Calculate the area of trapezoid .
10.Given that ABCD parallelogram with AB = 6 cm , AD = 4 cm and AC = 8 cm .
Calculate the area of the parallelogram .
===================================================================
Trigonometry / Class X . sem.2 / page - 43
Competence Check of Chapter 6
1. The value of :
X
sec(
13
 ). sin(  )  cot(
2
10
6
3
3
5
 ). cos ec(   )
3
7
4
2 tan(  ). cos(  )
2
3
A. – 4
B. – 2
C. – 1
D. – 3
E. 1
2.The solutions set √2 sin x – 1 = 0 for
- 6300  x  3600 is ….
A.{ 450 , 1350 }
B.{ 450 , 1350 , 2250 }
C.{ - 225 0 , 450 , 1350 }
D.{ - 2250 , - 450 , 1350 }
E.{ - 2250 , - 1350 , 450 }
3.The triangle ABC is right angled at C,
AC =6 cm, BC =8 cm and CD
perpendicular to AB then the length of
AD is …
A. 2,6 cm
D. 3 cm
B. 3,6 cm
E. 8 cm
C. 4,8 cm
4.In the triangle ABC given that cos
BAC = 9 , cos ACB = 3 , AB =12
4
16
cm and BC =15 cm then the length of
AC is …
A. 16 cm
D. 18 cm
B. 20 cm
E. 24 cm
C. 28 cm
5.The triangle WXY is right angled at W
and WXY =600, WW 1 and W 2W 3 are
perpendicular to XY and W1W 2 is
perpendicular to WY.
If XY =4 cm then WW 1 +W2W 3 =…
W1
W3
W
W2
3
A.
3 cm
4
B. 3 3 cm
2
C. 5 3 cm
2
Y
D. 1 3 cm
2
E. 5 3 cm
4
6.In the ABC, Ac =10 cm and AB =8
cm, if cos(b + c) = 9 then BC =…
40
A.8 2 cm
D.9 2 cm
B.10 2 cm
E.11 2 cm
C.12 2 cm
7.The maximum value of function y = 2
– 3 cos (2x + ) is …
A. 5
D.-1
B.-5
E.1
C.2
8.The solutions set 2 sin2x + sin x – 1 =
0 for -1800  x  3600 is ….
A.{-900, 300, 1500}
B.{300, 900, 1500}
C.{-900, -300, 1500}
D.{-1500, -900, 300}
E.{-900, 600, 1200}
9.Given that tan  = a, find cos2 –
sin2 = …
A. 1 – a2
D. 2(1 – a2)
1  a2
(1 a)2
B.
E.
2
(1  a)
1  a2
1  a2
C.
1  a2
Trigonometry / Class X . sem.2 / page - 44
10.Solving problems
:1
cosec2x  sec2x
=…
cosec2x  sec2x
A.2 sin2 x
B.2 sec2 x
C. –2 sin2 x
D.2 cos2 x
E.2 cosec2 x
11.In the ABC, AC =7 cm, AB =8 cm
and ABC =600 then cos C = …
A. 1 3
D. 1 3
7
7
4
13
7
1
C.
7
B.
E.
4
3
7
12.In the triangle ABC , AB = 4 cm , BC
= 6 cm and AC = 8 cm then the
value tan  ABC = …
1
15
7
B.  15
A.
15
7
15
D. 
15
8
15
E.
15
C.
17.If sin  = 3 and tan  = 7 ( ,  is
5
24
0
acute) then cos (360 –).sin (900 +)
+ cos (900 +).cos (2700 +) is …
A. 3
D. 5
4
3
3
4
B.
E.
5
5
5
C.
4
A. 2 1  x 2
B. – 2 1  x 2
C. – 2 x 2  1
14.On the  PQR with datum PR = 12
cm , Q = 1200 and  P =
300.Calculate area of  PQR =
…cm2.
A. 36 3
D. 24 3
C. 6 3
16.The solutions set 1+ tan2 x = 3 sec x
– 2 for - 3600 < x < 1800 is …
A.{-3600, -3000, -600, 00, 600}
B.{-3000, -600, 00, 600}
C.{-600, 00, 1200}
D.{-3000, -600, 00, 600}
E.{-3000, -2400, -600, 00, 600}
18.If cos 250 = x then sin 2050 – sin 1150
=…
13.In the triangle ABC , AC = 5 cm , AB
= 7 cm and  BCA = 1200 than
perimeter triangle is …
A. 14 cm
D. 15 cm
B. 16 cm
E. 17 cm
C. 18 cm
B. 18 3
15.In the triangle ABC, a =4 cm and b
=3cm. If the area of ABC = 8 cm2
then cos C = …
A. 1
D. 1 3
2
3
1
B.
E. 3
3
2
E.1
E. 12 3
D. 2 x 2  1
E. – 2 1  x 2
19.A ship sails 40 mil from S to T an a
bearing 0300 and then 60 mil from T
to U on a bearing 1500. Calculate
the distance SU = …mil.
A. 20 2
D. 20 3
B. 20 5
C. 20 11
E. 20 7
Trigonometry / Class X . sem.2 / page - 45
20.Two men begin to walk each from
point A to point B at same moment. In
order that they come to point C at the
same time , so the walking velocity of
the man from point A should be …
C
300
450
A
B
A. twice of the man’s velocity from B
B.
1
2
2 times the man’s velocity from B
2 the man’s velocity from B
D. 2 2 the man’s velocity from B
C.
3 the man’s velocity from B
E.
KEY
1
2
3
4
5
6
7
8
9
10
A
C
B
D
D
B
A
A
C
A
11
12
13
14
15
16
17
18
19
20
C
B
D
E
D
D
B
E
E
C
Trigonometry / Class X . sem.2 / page - 46
REFERENCES
1.H.H. Heng , Khoo Cheng and J.F.Talbert.2001 . Additional Mathematics ,
Singapore
2.Prof Lee Peng Yee , Dr Fan Liang Hou , The Keng seng Bsc, DipEd , Looi
Chin Keong Bsc , Dip Ed , New Syllabus Additional Mathematics , Sinapore
3.Suwah Sembiring , Cucun Cunayah , Ahmad Zaelani , Etsa Indra Irawan
,Matematika Bilingual ( KTSP ) Untuk SMA/MA Kelas X Sem.1 & 2 , YARAMA
WIDYA 9 2008 )
4. Sri Kurnianingsih , Kuntarti , Sulistiyono , Mathematics for Senior High School
Grade X 1 by esis an imprint of PT. Penerbit Erlangga 2009
5. Marwanta ,S,Pd , Drs. H. Sigit Supriojanto , Suwarsini Murniati , Drs.
Herynugraha , Drs. Kamta Agus Sajaka , Drs. H . Soetiyono ,Mathematics For
Senior Higgh School Year x 1 , Yudhistira 2008
6.Sukino , Matematika untuk kelas SMA Kelas X sem.2 , Penerbit Erlangga ,
2006
7.Sartono Wirodikromo , Matematika untuk SMA kelas X sem.2 jilid 2 , Erlangga ,
2004
8.Loedji , Willa , 2004 , Matematika Bilingual untuk SMA kelas X , sem.1 & 2 ,
Bandung : Yrama Widya
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