Question

From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 45° and 30° respectively. Find the height of the hill. [CBSE 2017]

Answer 1

Let PQ be the hill of height h km. Let R and S be two consecutive kilometre stones, so the distance between them is 1 km.
Let QR = x km.

$$\begin{gathered} \mathrm{In}∆\mathrm{PQR}, \\ \tan 45°=\frac{\mathrm{PQ}}{\mathrm{QR}} \\ \Rightarrow 1=\frac{h}{x} \\ \Rightarrow h=x...\left(i\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In}∆\mathrm{PQS}, \\ \tan 30°=\frac{\mathrm{PQ}}{\mathrm{QS}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+1} \\ \Rightarrow \sqrt{3}h=x+1...\left(\mathrm{ii}\right) \end{gathered}$$
From equation (i) and (ii) we get,
$$\begin{gathered} \sqrt{3}h=h+1 \\ \Rightarrow h\left(\sqrt{3}-1\right)=1 \\ \Rightarrow h=\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} \\ \Rightarrow h=\frac{\sqrt{3}+1}{2}=\frac{2.73}{2}=1.365\mathrm{km} \end{gathered}$$
Hence, the height of the hill is 1.365 km.