# 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 A
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 $\begin{array}{l} =\left[\frac{60(\sqrt{3}-1)}{5}\right] \mathrm{m} / \mathrm{s} \\ =(12 \times 0.73) \mathrm{m} / \mathrm{s} \\ =\left(12 \times 0.73 \times \frac{18}{5}\right) \mathrm{km} / \mathrm{hr} \\ =31.5 \mathrm{~km} / \mathrm{hr} \approx 32 \mathrm{~km} / \mathrm{hr} \end{array}$