A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 60° with the man's eye when at a distance of 60 metres from the tower. After 9 seconds, the angle of depression becomes 30°. The speed of the boat, assuming that it is running in still water, will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 48 kmph
The angle of depression from the man’s eye equals the angle of elevation of the man as seen from the boat, because the two horizontal lines are parallel. Each position of the boat then gives a right triangle with the tower, in which \(\tan\theta=\frac{\text{height}}{\text{distance}}\).
The height (first position):
\(\tan60^\circ=\displaystyle\frac{h}{60}\), so \(h=60\sqrt{3}\) m
The distance moved (after 9 s):
\(\tan30^\circ=\displaystyle\frac{h}{60+x}\)
So \(60+x=60\sqrt{3}\times\sqrt{3}=180\) m
\(x=120\) m
The speed (120 m in 9 s):
Speed \(=\displaystyle\frac{120}{9}=\frac{40}{3}\) m/s
In km/h: \(\displaystyle\frac{40}{3}\times\frac{18}{5}=48\) km/h
Shortcut:
The height is the same in both triangles, so the distances are in the ratio \(\tan60^\circ:\tan30^\circ=\sqrt{3}:\frac{1}{\sqrt{3}}=3:1\)
The second distance is \(3\times60=180\) m
So the boat covers 120 m in 9 s, that is 48 km/h