Question

The angles of elevation and depression of the top and bottom of a light-house from the top of a 60 m high building are 30° and 60° respectively. Find
(i) the difference between the heights of the light-house and the building.
(ii) the distance between the light-house and the building .

Answer 1

Let AE be the light house and BD be the building of 60 m height


In right triangle ABC
$$\begin{gathered} \tan 30°=\frac{\mathrm{AC}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} \\ \Rightarrow x=\sqrt{3}h \end{gathered}$$
Now, In right triangle BDE
$$\begin{gathered} \tan 60°=\frac{\mathrm{BD}}{\mathrm{DE}} \\ \Rightarrow \sqrt{3}=\frac{60}{x} \\ \Rightarrow x=\frac{60}{\sqrt{3}}=20\sqrt{3}\mathrm{m} \\ \mathrm{Now}, \\ x=\sqrt{3}h \\ \Rightarrow 20\sqrt{3}=\sqrt{3}h \\ \Rightarrow h=20\mathrm{m} \end{gathered}$$
(i) The difference between the heights of the light-house and the building is h = 20 m

(ii) The distance between the light-house and the building is x = $20\sqrt{3}$ m