Question

(i) A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.

(ii) A man in a boat rowing away from a light house 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

(i) ​Let the distance BC be x m and CD be y m.


$$\begin{gathered} \mathrm{In}∆\mathrm{ABC}, \\ \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{150}{x} \\ \Rightarrow \sqrt{3}=\frac{150}{x} \\ \Rightarrow x=\frac{150}{\sqrt{3}}\mathrm{m}.....\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In}∆\mathrm{ABD}, \\ \tan 45°=\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{150}{x+y} \\ \Rightarrow 1=\frac{150}{x+y} \\ \Rightarrow x+y=150 \\ \Rightarrow y=150-x \end{gathered}$$

Using (1), we get

$\Rightarrow y=150-\frac{150}{\sqrt{3}}=\frac{150\left(\sqrt{3}-1\right)}{\sqrt{3}}\mathrm{m}$
Time taken to move from point C to point D is 2 min = $\frac{2}{60}\mathrm{h}=\frac{1}{30}\mathrm{h}$.
Now,

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{y}{{\displaystyle \frac{1}{30}}}$
$=\frac{\frac{150\left(\sqrt{3}-1\right)}{\sqrt{3}}}{{\displaystyle \frac{1}{30}}}=1500\sqrt{3}\left(\sqrt{3}-1\right)\mathrm{m}/\mathrm{h}$

(ii)



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,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{100}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{100}{\sqrt{3}}=\frac{100\sqrt{3}}{3}\mathrm{m} \end{gathered}$$

In right ∆ABD,

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

Hence, the speed of the boat is 57.73 m/min.