Question

Two ships are there in the sea on either side of a light house in such away that the ships and the light house are in the same straight line. The angles of depression of two ships are observed from the top of the light house are 60º and 45º respectively. If the height of the light house is 200 m, find the distance between the two ships. (Use $\sqrt{3}$ = 1.73)

Answer 1

Let CD be the light house and A and B be the positions of the two ships.

Height of the light house, CD = 200 m

Now,

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

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

In right ∆ACD,

$$\begin{gathered} \tan 60°=\frac{\mathrm{CD}}{\mathrm{AC}} \\ \Rightarrow \sqrt{3}=\frac{200}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=\frac{200}{\sqrt{3}}=\frac{200\sqrt{3}}{3}\mathrm{m} \end{gathered}$$

In right ∆BCD,

$$\begin{gathered} \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{200}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=200\mathrm{m} \end{gathered}$$

∴ Distance between the two ships, AB = BC + AC
$$\begin{gathered} =200+\frac{200\sqrt{3}}{3} \\ =200+\frac{200\times 1.73}{3} \\ =200+115.33 \\ =315.33\mathrm{m}\left(\mathrm{approx}\right) \end{gathered}$$

Hence, the distance between the two ships is approximately 315.33 m.