A man is watching from the top of tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man’s eye when at a distance of 60 meters from the tower. After 5 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?
A. 32 kmph B. 36 kmph C. 38 kmph D. 40 kmph Answer: Option AShow Answer
Solution(By Apex Team)
Let AB be the tower and C and D be the two positions of the boats.
Then,
$\angle \mathrm{ACB}=45^{\circ}, \angle A D B=30^{\circ}$
and AC = 60 m
Let, AB = h
Then,
$\frac{A B}{A C}=\tan 45^{\circ}=1$
$\begin{array}{l}
\Rightarrow A B=A C \\
\Rightarrow h=60 m \\
\text { And } \frac{A B}{A D}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\
\Rightarrow A D=(A B \times \sqrt{3}) \\
\quad=60 \sqrt{3} m \\
\therefore C D=(A D-A C) \\
\quad=60(\sqrt{3}-1) m
\end{array}$
Hence, required speed
$\begin{array}{l}
=\left[\frac{60(\sqrt{3}-1)}{5}\right] \mathrm{m} / \mathrm{s} \\
=(12 \times 0.73) \mathrm{m} / \mathrm{s} \\
=\left(12 \times 0.73 \times \frac{18}{5}\right) \mathrm{km} / \mathrm{hr} \\
=31.5 \mathrm{~km} / \mathrm{hr} \approx 32 \mathrm{~km} / \mathrm{hr}
\end{array}$
