JEE Main 2018 (Online) 15th April Morning ShiftQuestion 1
An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt3$ km above it, is observed at an elevation of $60^{\circ}$ from a point on the ground. If, after five seconds, its elevation from the same point, is $30^{\circ}$, then the speed (in km/hr) of the aeroplane is
Trigonometric Ratio & Identities
Solution
Concept. The horizontal distance changes from $\sqrt3\cot60^\circ$ to $\sqrt3\cot30^\circ$. Formulas used. (1) $\text{distance}=h\cot(\text{elevation})$Step 1 - Distances. $\sqrt3\cdot\frac{1}{\sqrt3}=1$ km and $\sqrt3\cdot\sqrt3=3$ km. Step 2 - Speed. $2$ km in $5$ s: $\frac{2}{5/3600}=1440$ km/hr. Answer: (D)