Question

A man of height $6ft$ is moving away from a building of height $106ft$ at the rate of $25ft/sec.$ The absolute value of the rate at which the angle of elevation of the top of the building is changing, when he is at a distance of $50ft$ from the foot of the building is $rad/sec$

Answer 1

$AB=\text{Man},CD=\text{Building}$
Let $BC=xft$ $\Rightarrow \frac{dx}{dt}=25ft/sec$
$\tan \theta =\frac{100}{x}\Rightarrow {\sec }^2\theta \frac{d\theta }{dt}\cdot \frac{dt}{dx}=-\frac{100}{x^2}$
When $x=50ft$
$\tan \theta =\frac{100}{50}=2\Rightarrow \sec \theta =\sqrt{1+4}=\sqrt{5}$
Now,
$5\frac{d\theta }{dt}\cdot 25=-\frac{100}{{50}^2}\Rightarrow \frac{d\theta }{dt}=-2rad/s$

Hence the absolute value of the rate will be,
$|\frac{d\theta }{dt}|=2rad/s$