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, the angle of depression of the boat with the man’s eye is $30°$ (Ignore the 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 point $B$, where the angle of depression is$45°$.

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

$10(\sqrt{3}+1)$

Explanation for the correct option:

Step 1: Draw the figure according to the question:

Given, $\angle PAB=30°,\angle PBQ=45°$

From the figure, In $△AQP$ we have

$\begin{array}{rcl}\frac{h}{x+y}& =& \tan 30\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{h}{x+y}& =& \frac{1}{\sqrt{3}}\end{array}$

$\Rightarrow$$\begin{array}{rcl}x+y& =& \sqrt{3}h\dots .\dots \dots .\left(1\right)\end{array}$

Step 2: Finding the speed

In $△BQP$, we have

$\frac{h}{y}=\tan {45}^{0^{}}$

$\Rightarrow$$\frac{h}{y}=1$

$\Rightarrow$ $h=y\dots \dots .\left(2\right)$

put in (1)

$\begin{array}{rcl}x+y& =& \sqrt{3}y\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& (\sqrt{3}–1)y\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{x}{20}& =& V\end{array}$ (speed)

Step 3: Time taken to reach foot from $B$

Time taken$\begin{array}{rcl}& =& \frac{y}{V}\end{array}$

$\begin{array}{rcl}& =& \frac{20x}{(\sqrt{3}-1)x}\\ & =& \frac{20(\sqrt{3}+1)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ & =& \frac{20(\sqrt{3}+1)}{2}\\ & =& 10(\sqrt{3}+1)\end{array}$

$\therefore$The time taken (in seconds) by the boat from $B$ to reach the base of the tower is $\mathbf{10}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{)}$

Hence, The correct answer is option (D).