question_answer

A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point 0 on the ground is \[45{}^\circ \]. It flies off horizontally straight away from the point 0. After one second, the elevation of the bird from 0 is reduced to \[30{}^\circ \]. Then the speed (in m/s) of the bird is

A) \[40\left( \sqrt{2}-1 \right)\]

B) \[40\left( \sqrt{3}-\sqrt{2} \right)\]

C) \[20\sqrt{2}\]

D) \[20\left( \sqrt{3}-1 \right)\]

Correct Answer: D

Solution :

\[\tan 30{}^\circ =\frac{20}{20+x}=\frac{1}{\sqrt{3}}\]

\[20x=20\sqrt{3}\]

\[x=20\left( \sqrt{3}-1 \right)\]

\[\Rightarrow \] Speed is \[20\left( \sqrt{3}-1 \right)\text{m/sec}\]