Question

From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45°, respectively. Find the distance between the cars. [Take $\sqrt{3}$= 1.73] [CBSE 2011]

Answer 1

Let PQ be the tower.

We have,

$$\begin{gathered} \mathrm{PQ}=100\mathrm{m},\angle \mathrm{PAQ}=30°\mathrm{and}\angle \mathrm{PBQ}=45° \\ \mathrm{In}∆\mathrm{APQ}, \\ \tan 30°=\frac{\mathrm{PQ}}{\mathrm{AP}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{\mathrm{AP}} \\ \Rightarrow \mathrm{AP}=100\sqrt{3}\mathrm{m} \\ \mathrm{Als},\mathrm{in}∆\mathrm{BPQ}, \\ \tan 45°=\frac{\mathrm{PQ}}{\mathrm{BP}} \\ \Rightarrow 1=\frac{100}{\mathrm{BP}} \\ \Rightarrow \mathrm{BP}=100\mathrm{m} \\ \mathrm{Now},\mathrm{AB}=\mathrm{AP}+\mathrm{BP} \\ =100\sqrt{3}+100 \\ =100\left(\sqrt{3}+1\right) \\ =100\times \left(1.73+1\right) \\ =100\times 2.73 \\ =273\mathrm{m} \end{gathered}$$

So, the distance between the cars is 273 m.