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

Answer 1

The correct option is D $10(\sqrt{3}+1)$


$\frac{h}{x+y}=\tan {30}^{\circ }$
$x+y=\sqrt{3}h$ $....\left(1\right)$
also
$\frac{h}{y}=\tan {45}^{\circ }$
$h=y$ $...\left(2\right)$
put $h=y$ in equation $\left(1\right)$
$x+y=\sqrt{3}y$
$x=(\sqrt{3}-1)y$
and
$\frac{x}{20}={}^{\prime }v{}^{\prime }$ $(v=$ speed$)$
$\therefore$ time taken to reach foot from $B$
$=\frac{y}{v}$
$=\frac{x}{(\sqrt{3}-1)\times x}\times 20$
$=10(\sqrt{3}+1)$