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 how much more time it will take to reach the shore.

Answer 1

Let height of cliff= $AB=h$

$\angle EAD=\angle ADB$ (alternate interior $\angle$)

$={30}^o$ (Refer to Image)

Also, $\angle EAC=\angle ACB={60}^o$

Let speed of boat= $V$ m/s

$A/Q$, $DC=3\times 60V=180V$

Let, $BC=x$

Time taken from $C$ to $B$=$t$

$\Rightarrow x=vt$

Now, in ${△}^{le}ABD,\tan {30}^o=\frac{h}{180v+vt}$

$\Rightarrow 1\sqrt{3}=\frac{h}{180v+vt}$

In ${△}^{le}ABC$, $\tan {60}^o=\frac{h}{vt}$

$\Rightarrow \sqrt{3}=\frac{h}{vt}$

$\Rightarrow h=\sqrt{3}vt$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\sqrt{3}vt}{180v+vt}$

$\Rightarrow 180+t=3t$

$\Rightarrow 180=2t\Rightarrow t=90s$= $1min30sec$