Question

A man is moving away from a tower 41.6m high at a rate of 2 m/s. If the eye level of the man is 1.6m above the ground, then the rate at which the angle of elevation of the top of the tower is changing when he is at a distance of 30 m from the foot of the tower is

• A. $-\frac{4}{125}radian/sec$
• B. $-\frac{2}{25}radian/sec$
• C. $-\frac{1}{625}radian/sec$
• D. None of these

Answer 1

The correct option is A

$-\frac{4}{125}radian/sec$

Let CD be the position of man at any time t. Let BD = x, then EC = x. Let $\angle ACE=\theta$
Given, AB = 41.6 m, CD = 1.6 m and $\frac{dx}{dt}=2m/sec$
AE = AB - EB = AB - CD = 41.6 - 1.6 = 40m
We have to find $\frac{d\theta }{dt}$ when x = 30 m
From $\Delta AEC,tan\theta =\frac{AE}{EC}=\frac{40}{x}$ . . . (1)
Diff. w.r.t to t,
$se{c}^2\theta \frac{d\theta }{dt}=\frac{-80}{x^2}co{s}^2\theta =-\frac{80}{x^2}.\frac{x^2}{x^2+{40}^2}[\because cos\theta =\frac{x}{\sqrt{x^2+{40}^2}}]\Rightarrow \frac{d\theta }{dt}=-\frac{80}{x^2-{40}^2}\cdots \left(2\right)whenx=30m,\frac{d\theta }{dt}=-\frac{80}{{30}^2+{40}^2}=-\frac{4}{125}radian/sec$
Hence the correct option is (a)