Question

On the level ground, the angle of elevation of a tower is 30°. On moving 20 meters nearer, the angle of elevation is 60°. The height of the tower is ____________.

Answer 1

Let AB be the tower of height h m. Suppose C and D be the positions on the level ground such that CD = 20 m.



In right ∆ABC,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{h}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{h}{\sqrt{3}} \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD}=\sqrt{3}h \end{gathered}$$

Now,

BD − BC = 20 m

$$\begin{gathered} \Rightarrow \sqrt{3}h-\frac{h}{\sqrt{3}}=20 \\ \Rightarrow \frac{3h-h}{\sqrt{3}}=20 \\ \Rightarrow 2h=20\sqrt{3} \\ \Rightarrow h=10\sqrt{3}\mathrm{m} \end{gathered}$$

Hence, the height of the tower is $10\sqrt{3}\mathrm{m}$.

On the level ground, the angle of elevation of a tower is 30°. On moving 20 meters nearer, the angle of elevation is 60°. The height of the tower is $\underline{10\sqrt{3}\mathrm{m}}$.