Question

A man on a cliff observes a boat, at an angle of depression ${30}^{\circ }$, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be ${60}^{\circ }$. Assuming that the boat sails at a uniform speed, determine :

(i) how much more time it will take to reach the shore ?

(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.

Answer 1

Let the cliff be $AB=500m$ and the speed of the boat be $xm/sec$

Distance covered in 3 min $=CD=3\times 60\times x=180xm$
(as distance $=$ speed $\times$time)

In ∆ABC,
$\tan {60}^{\circ }=\frac{AB}{BC}=\frac{500}{BC}\Rightarrow BC=\frac{500}{\sqrt{3}}$

In ∆ABD, $\tan {30}^{\circ }=\frac{AB}{BD}=\frac{500}{BD}\Rightarrow BD=500\times \sqrt{3}\therefore BD=BC+CD\Rightarrow 500\sqrt{3}=\frac{500}{\sqrt{3}}+180x\Rightarrow x=3.2$

So, Speed of the boat is $3.2m/sec$

Distance to be covered $=BC$

[$\because$ Time $=\frac{\text{Distance}}{\text{Speed}}$]

Time taken to cover BC $=\frac{\frac{500}{\sqrt{3}}}{3.2}=90secs=1.5min$