Question

An aircraft is flying at a uniform speed $vm{s}^{-1}$. If the angle substend at an observation point on the ground by two positions of the aircraft $t$ seconds apart is $\theta$, the height of the aircraft above the ground is given by :

• A. $\frac{vt}{2tan\theta }$
• B. $\frac{2vt}{tan\theta }$
• C. $\frac{vt}{tan\left[\frac{\theta }{2}\right]}$
• D. $\frac{vt}{2tan\left[\frac{\theta }{2}\right]}$

Answer 1

Answer: (D)

$\frac{vt}{2tan\left[\frac{\theta }{2}\right]}$

Please refer to following image

Let P, Q, and R denotes the position of the aircraft as it flies above the observational point O and the angle subtend here is $\theta$ from end to end.

and, $OR$ = height of aircraft from ground level $=h$

Using trignometry in triangle PQO,
So, by applying tan $\theta /2$ = (perpendicular/base)
$tan\left(\theta /2\right)=\frac{vt/2}{h}$

$\therefore$ $h=\frac{vt}{2\tan (\theta /2)}$