Question

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 the tower is

• A. 10
• B. $10\sqrt{3}$
• C. $10(\sqrt{3}+1)$
• D. $10(\sqrt{3}-1)$

Answer 1

Answer: (C)

Let speed of boat is $um/s$ and height of tower is $h$ meter & distance $AB=x$ metre
$\therefore x=h\cot 30^{\circ }-h\cot 45^{\circ }$
$\Rightarrow x=h(\sqrt{3}-1)$
$\therefore u=\frac{x}{20}=\frac{h(\sqrt{3}-1)}{20}m/s$
$\therefore$ Time taken to travel from $B$ to $C$ (Distance $=h$ meter
$=\frac{h}{u}=\frac{h}{h\frac{(\sqrt{3}-1)}{20}}=\frac{20}{\sqrt{3}-1}=10(\sqrt{3}+1)sec$