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 lighthouse from 60° to 30°. Show that the speed of the boat is $\frac{100\sqrt{3}}{3}$ metres per minute.

Answer 1

Let $AB$ be the light house and $C$ and $D$ be the positions of the two ships. Thus, we have:
∠$ACB=30°$
∠$ADB={60}^o$
$AB=100$ m
Let:
$BC\hspace{0.17em}=x$ m and $CD\hspace{0.17em}=y$ m
Thus, we have:
$BD=(BC-CD)=(x-y)$ m

In ∆$ABC$, we have:

$$\begin{gathered} \frac{AB}{BC}=\tan {30}^o=\frac{1}{\sqrt{3}} \\ \Rightarrow \frac{100}{x}=\frac{1}{\sqrt{3}} \\ \Rightarrow x=100\sqrt{3} \end{gathered}$$

In ∆$ABD$, we have:

$$\begin{gathered} \frac{AB}{BD}=\tan {60}^o=\sqrt{3} \\ \Rightarrow \frac{100}{(x-y)}=\sqrt{3} \end{gathered}$$
$\Rightarrow$ $(x-y)=\frac{100}{\sqrt{3}}$
∴ $y=x-\frac{100}{\sqrt{3}}=100\sqrt{3}-\frac{100}{\sqrt{3}}=\frac{300-100}{\sqrt{3}}=\frac{200}{\sqrt{3}}$ m
​Distance travelled in two minutes = $CD=y=\frac{200}{\sqrt{3}}$ = $\frac{200\sqrt{3}}{3}$ m
$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{{\displaystyle \frac{200\sqrt{3}}{3}}}{2}=\frac{100\sqrt{3}}{3}$ metres per minute

Hence proved.