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 $C$ be the cliff and $A$ and $D$ be the two positions of the boat.
We have:
∠$BAC={30}^o$ and ∠$BDC={60}^o$
Let the speed of the boat be $v$ metres per minute.
Also,
$AD=$ Distance covered by the boat in six minutes
$=6v$ m
Suppose the boat takes $t$ minutes to reach $B$ from $D$.
∴ $DB=vt$ m
Let:
$BC=h$ m

In the right ∆$DBC$, we have:
$\frac{BC}{DC}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{h}{vt}=\sqrt{3}$
⇒ $h=\sqrt{3}vt$ ...(i)

From the right ∆$ACB$, we have:
$\frac{BC}{AC}=\tan {30}^o$ $=\frac{1}{\sqrt{3}}$
⇒ $\frac{h}{(6v+vt)}=\frac{1}{\sqrt{3}}$

⇒ $h=\frac{(6v+vt)}{\sqrt{3}}$ ...(ii)
From (i) and (ii), we have:
$\sqrt{3}vt=\frac{(6v+vt)}{\sqrt{3}}$
⇒ $3vt=6v+vt$
⇒ $2vt=6v$
$t=\frac{6v}{2v}=3$ minutes
Hence, the required time is three minutes.