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?

Answer: Option A

Solution

Answer: Option A

Solution:

Let AB be the tower and C and D be the two positions of the boats.
Then, $$\angle \text{ACB = }45^{\circ },$$    $$\angle ADB=30^{\circ }$$     and AC = 60 m
Let, AB = h
$$\begin{array}{rl}& \text{Then,}\frac{AB}{AC}=\tan 45^{\circ }=1\\ & \Rightarrow AB=AC\\ & \Rightarrow h=60m\\ & \text{And}\frac{AB}{AD}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}\\ & \Rightarrow AD=\left(AB\times \sqrt{3}\right)\\ & =60\sqrt{3}m\\ & \therefore CD=\left(AD-AC\right)\\ & =60\left(\sqrt{3}-1\right)m\\ & \text{Hence, required speed }\\ & \text{ = }\left[\frac{60\left(\sqrt{3}-1\right)}{5}\right]\text{m/s}\\ & \text{ = }\left(12\times 0.73\right)\text{m/s}\\ & =\left(12\times 0.73\times \frac{18}{5}\right)\text{km/hr}\\ & =\text{31}\text{.5km/hr }\approx \text{ 32km/hr}\end{array}$$