Question

A man on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60°. Find the time taken by the boat to reach the shore.

Answer 1

Let PC be the cliff and A the initial position of the boat, where the angle of depression is 30°. The boat reaches point B after 6 minutes, where the angle of depression is 60°. 

In Δs APC and BPC, 

tan 30° = \(\frac{PC}{AC}\) and tan 60° = \(\frac{PC}{BC}\) ⇒ \(\frac{1}{\sqrt3}\) = \(\frac{PC}{AC}\) and √3 = \(\frac{PC}{BC}\)

⇒ PC = \(\frac{AC}{\sqrt3}\) and PC = √3 BC ⇒ \(\frac{AC}{\sqrt3}\) = √3 BC ⇒ AC = 3BC.

Now AB = AC – BC = AC \(-\frac{1}{3}\) AC = \(\frac{2}{3}\) AC ⇒ AB = \(\frac{2}{3}\) AC

Let the speed of the boat be x metres/minutes. Then 

AB = Distance travelled by the boat in 6 minutes = 6\(x\)

From eq. (i), we have \(\frac{2}{3}\) AC = 6\(x\) ⇒ AC = 9\(x\)

∴ Time taken by the boat to reach the shore = \(\frac{AC}{x}\)       [T = \(\frac{D}{S}\)]

= \(\frac{9x}{x}\) minutes = 9 minutes.