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TRIGONOMETRY
KALPALATHA
GHS
SANTHIPURAM
SOGADABALLA
NAME OF THE SIEDES OF RIGHTANGLE TRIANGLE
OPPOSITESIDE
C
A
ADJESENTSIDE
B
θ
A
ADJESENTSIDE
OPPOSITESIDE
Sinθ=
θ
A
ADJESENTSIDE
OPPOSITESIDE
Cosθ=
θ
A
ADJESENTSIDE
OPPOSITESIDE
Cosecθ=
θ
A
ADJESENTSIDE
OPPOSITESIDE
Secθ=
θ
ADJESENTSIDE
OPPOSITESIDE
Tanθ=
θ
ADJESENTSIDE
OPPOSITESIDE
Cotθ=
Sinθ
0º
30º
45º
60º
90º
0
1/2
1/√2 √3/2
0
1/√3
1
√3
∞
2
√2
2/√3
1
1
Cosθ
Tanθ
Cscθ ∞
Secθ
Cotθ
OBJECT
ANGLE OF ELEVATION
OBSERVAR
HORIZONTAL
ANGLE OF DEPRESSION
HORIZANTAL
Angle of depression
HORIZANTAL
OBJECT
There are two temples, one on each
bank of the river, just opposite to each
Other .one of the temples A is 40 mts
high .As observed from the top of this
temple A , the angle of depression of
the top and foot of the other temple B
are 12°.30' and 21°.48' respectively.
Find the width of the river and the
height of the temple B approximately.
12°.30'
21°.48'
12°.30'
21°.48'
90°
AB= river
C
AC=BD=temples
C=observer
In right angle triangle
ABC
E
TanB=AC/AB
12°.30'
D
Tan 21°.48' =40/AB
0.400
=40/AB
AB
=40/0.4
AB
=100 mts
In right angle triangle
21°.48'
A
CED
TanD=CE/ED
Tan12°.30‘=CE/100
CE =0.2217×100=22.17 mts
Height of the temple AC=AE+EC
40 =BD+22.17
( BD=AE)
BD =40-22.17=17.83 mts
Height of the B temple =17.83 mts
B
An aeroplane at an altitude of 2500
mts . Observes the angles of
depression of opposite points on
the two banks of river to be 41°.20'
and 52°.10' . Find the width of the
river in metres.
41°.20'
41°.20'
52°.10'
52°.10'
AB=2500mts
A
C and D are two bank river.
2500mts
CD is the width of the river
Given ∟ACB= 41°.20‘ and ∟ADB= 52°.10‘
From right angle triangle ACB
tan 41°.20‘ =AB/CB
0.8795=2500/CB
C
CB = 2500/0.8795
= 2842.5mts
from right angle triangle ABD, ∟ADB = 52°.10‘
Tan 52°.10‘ = AB/BD
1.2876 = 2500/BD
BD =2500/1.2876=1941.5mts
CD=CB+BD
=2842.5+1941.5=4784 mts
The width of the river =4784 mts
B
E
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