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UPSC NDA Math Model Paper 4
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© examsnet.com
Question : 14
Total: 120
From three collinear points A, B and C on a level ground, which are on the same side of a tower, the angles of elevation of the top of the tower are 30°, 45° and 60° respectively. If BC = 60 m, then AB is:
15
√
3
m
30
√
3
m
45
√
3
m
60
√
3
m
Validate
Solution:
It can be noted that the angle of elevation from closer points is more than the angle from farther points.
The position of the three points is shown in the following diagram:
Using the definition of trigonometric ratio tan θ:
A
P
=
Q
P
tan
30
°
=
√
3
Q
P
B
P
=
Q
P
tan
45
°
=
Q
P
C
P
=
Q
P
tan
60
°
=
Q
P
√
3
It is given that
B
C
=
60
m
.
⇒
B
P
−
C
P
=
60
⇒
Q
P
−
Q
P
√
3
=
60
⇒
Q
P
=
60
√
3
√
3
−
1
Now
,
A
B
=
A
P
−
B
P
=
√
3
Q
P
−
Q
P
=
Q
P
(
3
−
1
)
=
60
√
3
√
3
−
1
(
√
3
−
1
)
=
60
√
3
m
© examsnet.com
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