Question

A boat is being rowed away from a cliff $150$ m high. From the top of the cliff, the angle of depression of the boat changes from 60${}^o$ to 45${}^o$ in 1 minute. If $\sqrt{3}=1.73$, the speed of the boat is _____

• A. $4.31$ km/hr
• B. $3.81$ km/hr
• C. $4.63$ km/hr
• D. $3.91$ km/hr

Answer 1

The correct option is B $3.81$ km/hr

Referring to the figure, let AB be the cliff and C and D be the two positions of the boat.

Then, $AB=150$ m, $\angle ACB={60}^o$ and $\angle ADB={45}^o$

Now, $cot{45}^o$ = $\frac{AD}{AB}\Rightarrow 1=\frac{AD}{150}\Rightarrow$ AD = 150 m

and $cot{60}^o$ = $\frac{AC}{AB}\Rightarrow \frac{1}{\sqrt{3}}=\frac{AC}{150}\Rightarrow AC=\frac{150}{\sqrt{3}}$ m

$\therefore$ AC = $\frac{150\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$ = 50$\sqrt{3}$ = 50 x 1.73 = 86.5 m.

$\therefore CD=AD-AC=(150-86.5)=63.5$ m

$\therefore$ Speed of the boat = $\frac{63.5}{60}$ m/s

= $(\frac{63.5}{60}\times \frac{18}{5})$ km/her

= $3.81$ km/hr

So, option B is correct.