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 $u m / s$ 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
$=\frac{ h }{ u }=\frac{ h }{ h \frac{(\sqrt{3}-1)}{20}}=\frac{20}{\sqrt{3}-1}=10(\sqrt{3}+1) sec$