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

Solution :
Let $$CD$$  be the position of man at any time $$t.$$
Let $$BD$$  be $$x.$$
Then $$EC = x.$$
Let $$\angle ACE$$  be $$\theta $$
Given $$AB = 41.6\,m,\,CD = 1.6\,m,{\text{ and }}\frac{{dx}}{{dt}} = 2{\text{ }}m/s.$$

$$AE = AB - EB = AB - CD = 41.6 - 1.6 = 40\,m$$
We have to find $$\frac{{d\theta }}{{dt}}$$  when $$x = 30\,m$$
From $$\Delta AEC,\,\tan \,\theta = \frac{{AE}}{{EC}} = \frac{{40}}{x}$$
Differentiating w.r.t. to $$t,$$
$$\eqalign{ & {\sec ^2}\theta \frac{{d\theta }}{{dt}} = \frac{{ - 40}}{{{x^2}}}\frac{{dx}}{{dt}} \cr & {\text{or }}{\sec ^2}\theta \frac{{d\theta }}{{dt}} = \frac{{ - 40}}{{{x^2}}} \times 2 \cr & {\text{or }}\frac{{d\theta }}{{dt}} = \frac{{ - 80}}{{{x^2}}}{\cos ^2}\theta \cr & {\text{or }}\frac{{d\theta }}{{dt}} = - \frac{{80}}{{{x^2}}}\frac{{{x^2}}}{{{x^2} + {{40}^2}}} \cr & {\text{or }}\frac{{d\theta }}{{dt}} = - \frac{{80}}{{{x^2} + {{40}^2}}} \cr} $$
When $$x = 30\,m,\,\,\,\frac{{d\theta }}{{dt}} = - \frac{{80}}{{{{30}^2} + {{40}^2}}} = - \frac{4}{{125}}\,rad/s.$$