Question

A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60° to 30°. Find the speed of the boat in metres per minute. [Use $\sqrt{3}$ = 1.732.]

Answer 1

Let AB be the light house. Suppose C and D be the two positions of the boat.

Here, AB = 100 m.

Let the speed of the boat be v m/min.

So,

CD = v m/min × 2 min = 2v m [Distance = Speed × Time]

In right ∆ABC,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{100\mathrm{m}}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{100}{\sqrt{3}}=\frac{100\sqrt{3}}{3}\mathrm{m}.....\left(1\right) \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{100\mathrm{m}}{\mathrm{BC}+\mathrm{CD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{{\displaystyle \frac{100\sqrt{3}}{3}}+2v}\left[\mathrm{From}\left(1\right)\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow 2v+\frac{100\sqrt{3}}{3}=100\sqrt{3} \\ \Rightarrow 2v=100\sqrt{3}-\frac{100\sqrt{3}}{3} \\ \Rightarrow 2v=100\sqrt{3}\left(1-\frac{1}{3}\right) \\ \Rightarrow 2v=100\sqrt{3}\times \frac{2}{3} \end{gathered}$$
$\Rightarrow v=\frac{100\times 1.732}{3}=57.73\mathrm{m}/\min$

Thus, the speed boat is 57.73 m/min.