SSC CGL 3 Sep 2016 Paper

Question : 68 of 100

Marks: +1, -0

From a point P on a level ground, the angle of elevation to the top of the tower is 30°. If the tower is 100 metre high, the distance of point P from the foot of the tower is
(Take

√3

= 1.73)

149 metre
156 metre
173 metre
188 metre

Validate

Solution:

Let, AB = Height of tower = 100 metre
∠ACB = 30°
In ΔABC, tan 30° =

{AB}/{BC}

⇒

1/√3

{100}/{BC}

⇒ BC = 100

√3

metre
= (100 × 1.73) metre
= 173 metre

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