JEE Main 2014 (Offline)Question 1
A bird is sitting on the top of a vertical pole $20$ m 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 :
Trigonometric Ratio & Identities
Solution
Concept. The bird stays at height $20$ m, so only its horizontal distance from $O$ changes. Formulas used. (1) Horizontal distance $=20\cot\theta$ for elevation $\theta$Step 1 - Before. $20\cot45^\circ=20$ m. Step 2 - After one second. $20\cot30^\circ=20\sqrt3$ m. Step 3 - Speed. Distance flown in $1$ s: $20\sqrt3-20=20(\sqrt3-1)$ m/s. Answer: (B)