Question

An aeroplane is flying at a height of 210 m. Flying at this height at some instant the angles of depression of two points in opposite directions on both the banks of the river are 45º and 60º. Find the width of the river. (Use $\sqrt{3}$ = 1.73) [CBSE 2015]

Answer 1

Let CD be the height of the aeroplane above the river at some instant and A and B be the two points on the opposite banks of the river.

Height of the aeroplane above the river, CD = 210 m

Now,

$\angle$CAD = $\angle$ADX = 60º (Alternate angles)

$\angle$CBD = $\angle$BDY = 45º (Alternate angles)

In right ∆ACD,

$$\begin{gathered} \tan 60°=\frac{\mathrm{CD}}{\mathrm{AC}} \\ \Rightarrow \sqrt{3}=\frac{210}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=\frac{210}{\sqrt{3}}=70\sqrt{3}\mathrm{m} \end{gathered}$$

In right ∆BCD,

$$\begin{gathered} \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{210}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=210\mathrm{m} \end{gathered}$$

∴ Width of the river, AB = BC + AC
$$\begin{gathered} =210+70\sqrt{3} \\ =210+70\times 1.73 \\ =210+121.1 \\ =331.1\mathrm{m} \end{gathered}$$

Hence, the width of the river is 331.1 m.