Question

A man on a cliff observes a boat at an angle of depression ${30}^{\circ }$, which is sailing towards the point of the shore 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, time it will take to reach the shore is 1 minute and 30 seconds
If true then enter $1$ and if false then enter $0$

Answer 1

Let the height of the cliff = $h$
Let the distance between the boat and the cliff, when it makes an angle of ${30}^{\circ }$ = $d_a$
Let the distance between the boat and the cliff, hen it makes an angle of ${60}^{\circ }$ = $d_b$
Distance traveled in $90$ sec = $d_a-d_b$
The ground, cliff and line of sight forms a right angled triangle,
Thus, $\tan \angle$ of elevation = $\frac{height}{distance}$
Now, $\tan 30=\frac{h}{d_a}$
$d_a=h\sqrt{3}$
and, $\tan 60=\frac{h}{d_b}$
$d_b=\frac{h}{\sqrt{3}}$
Speed of the boat = $\frac{distance}{time}=\frac{h\sqrt{3}-\frac{h}{\sqrt{3}}}{180}$
= $\frac{2h}{180\sqrt{3}}$
= $\frac{h}{90\sqrt{3}}$
Now, time taken to cover the distance, $d_a$
time = $\frac{d_a}{speed}$ = $\frac{\frac{h}{\sqrt{3}}}{\frac{h}{90\sqrt{3}}}$ = $90$ sec