JEE Main 2021 (Online) 25th February Morning ShiftQuestion 1
A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point $A$, with uniform speed. At that point, angle of depression of the boat with the man's eye is $30^{\circ}$ (Ignore man's height). After sailing for $20$ seconds, towards the base of the tower (which is at the level of water), the boat has reached a point $B$, where the angle of depression is $45^{\circ}$. Then the time taken (in seconds) by the boat from $B$ to reach the base of tower is :
Trigonometric Ratio & Identities
Solution
Concept. Find both distances from the base in terms of the height $h$; the speed is the same throughout. Formulas used. (1) $\text{distance}=h\cot(\text{depression})$Step 1 - Distances. By (1), $OA=\sqrt3h$ and $OB=h$, so $AB=(\sqrt3-1)h$. Step 2 - Speed. $AB$ takes $20$ s, so the speed is $\dfrac{(\sqrt3-1)h}{20}$. Step 3 - Time for $BO$. $$\frac{h}{(\sqrt3-1)h/20}=\frac{20}{\sqrt3-1}=10(\sqrt3+1).$$ Answer: (B)