Question

A man is watching from the top of a tower, a boat speeding away from the tower. The boat makes an angle of depression of ${45}^0$ with the man's eye when at a distance of 60 meters from the tower. After 5 seconds the angle of depression becomes ${30}^0$. What is the approximate speed of the boat assuming that it is running in still water?

• A. $32$ kmph
• B. $36$ kmph
• C. $38$ kmph
• D. $40$ kmph
• E. $42$ kmph

Answer 1

The correct option is A $32$ kmph

Let $AB$ be the tower and $C$ and $D$ be the point of observation from boat.

Given $AC=60\text{m}$ and $\angle ADB={30}^0$ and $\angle ACB={45}^0$

Let height of tower is $h\text{m}$

In $\Delta BAC$,

$\tan {45}^0=\frac{AB}{AC}$

$\Rightarrow 1=\frac{AB}{60}$

$\Rightarrow AB=60\text{m}$

In $\Delta BAD$,

$\tan {30}^o=\frac{AB}{AD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{60}{AD}$

$\Rightarrow AD=60\sqrt{3}\text{m}$

$\therefore CD=AD-AC=60\sqrt{3}-60=60(\sqrt{3}-1)$

Then, speed $=\frac{60(\sqrt{3}-1)}{5}\text{m/s}$

And approximate speed $=\frac{12\times 0.73\times 3600}{1000}\approx 32\text{km/h}$