AP ICET 2017 Shift 2 Paper

Question : 121 of 200

Marks: +1, -0

The angle of elevation o f the top of a tower from a point P on the ground is 45° and moving s- feet towards the tower the angle of elevation is found to be 60°. The height (in feet) of the tower is

$√3s$
$(√3-1)s$
$(√3+1)s$
$3/{√3-1}s$

Validate

Solution:

\tan60^{∘}={h}/{OQ}⇒√{3}={h}/{OQ}⇒OQ={h}/{√{3}}

OP = OT

[∠P=45°]

⇒s+{OQ}=h⇒s+h/{√{3}}=h⇒h(1-{1}/{√{3}})=s

∴h={√{3}}/{(√{3}-1)}s

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