Question

From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower. [CBSE 2011]

Answer 1

Let OP be the tower and points A and B be the positions of the cars.

We have,

$$\begin{gathered} \mathrm{AB}=100\mathrm{m},\angle \mathrm{OAP}=60°\mathrm{and}\angle \mathrm{OBP}=45° \\ \mathrm{Let}\mathrm{OP}=h \\ \mathrm{In}∆\mathrm{AOP}, \\ \tan 60°=\frac{\mathrm{OP}}{\mathrm{OA}} \\ \Rightarrow \sqrt{3}=\frac{h}{\mathrm{OA}} \\ \Rightarrow \mathrm{OA}=\frac{h}{\sqrt{3}} \\ \mathrm{Also},\mathrm{in}∆\mathrm{BOP}, \\ \tan 45°=\frac{\mathrm{OP}}{\mathrm{OB}} \\ \Rightarrow 1=\frac{h}{\mathrm{OB}} \\ \Rightarrow \mathrm{OB}=h \\ \mathrm{Now},\mathrm{OB}-\mathrm{OA}=100 \\ \Rightarrow h-\frac{h}{\sqrt{3}}=100 \\ \Rightarrow \frac{h\sqrt{3}-h}{\sqrt{3}}=100 \\ \Rightarrow \frac{h\left(\sqrt{3}-1\right)}{\sqrt{3}}=100 \\ \Rightarrow h=\frac{100\sqrt{3}}{\left(\sqrt{3}-1\right)}\times \frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)} \\ \Rightarrow h=\frac{100\sqrt{3}\left(\sqrt{3}+1\right)}{\left(3-1\right)} \\ \Rightarrow h=\frac{100\left(3+\sqrt{3}\right)}{2} \\ \Rightarrow h=50\left(3+1.732\right) \\ \Rightarrow h=50\left(4.732\right) \\ \therefore h=236.6\mathrm{m} \end{gathered}$$

So, the height of the tower is 236.6 m.

Disclaimer: The answer given in the texbook is incorrect. The same has been rectified above.