Question

A bird is sitting on the top of a vertical pole $20m$ high and its elevation from a point $O$ on the ground is ${45}^{\circ }$. It flies off horizontally straight away from the point $O$. After one second, the elevation of the bird from $O$ is reduced to ${30}^{\circ }$. Then the speed (in $m/s$) of the bird is

• A. $40(\sqrt{2}-1)$
• B. $40(\sqrt{3}-\sqrt{2})$
• C. $20\sqrt{2}$
• D. $20(\sqrt{3}-1)$

Answer 1

The correct option is D $20(\sqrt{3}-1)$

$\angle BAC={45}^{\circ },\angle DAE={30}^{\circ }\text{and}BC=DE=20m$


$AB=20\tan {45}^{\circ }=20$ and $AD=20\cot {30}^{\circ }=20\sqrt{3}$ m
$BD=AD-AB=20\sqrt{3}-20=20(\sqrt{3}-1)$
$\therefore$ speed of bird $=20(\sqrt{3}-1)m/s$