Question

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds , its elevation is observed to be 30°. Find the speed of the aeroplane in km /hr.

Answer 1

$tan60= \frac{P}{B_{1}}$
$\sqrt{3}= \frac{1}{B_{1}}$
$B_{1}= \frac{1}{ \sqrt{3} }$

$tan30= \frac{P}{B_{2}}$
$\sqrt{3} = \frac{1}{B_{2}}$
$B_{2}= \sqrt{3}$

Distance travelled by plane = $B_{2}-B_{1}$ = $\sqrt{3} - \frac{1}{ \sqrt{3}} = \frac{2}{ \sqrt{3}}$

Speed of plane (in km/h) = $\frac{Distance Travelled}{Time Taken} *60*60$
Speed = $\frac{ \frac{2}{ \sqrt{3}}}{10} *360$
Speed = 41.56km/s