Question

If the elevation of the sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high is _________.

Answer 1

Let AB be the pole. Suppose BD and BC be the lengths of the shadow of the pole when the elevation of Sun is 30° and 60°, respectively.



Here, AB = 15 m

In right ∆ABC,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{15}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{15}{\sqrt{3}}=5\sqrt{3}\mathrm{m}.....\left(1\right) \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{15}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD}=15\sqrt{3}\mathrm{m}.....\left(2\right) \end{gathered}$$

Now,

Difference between the length of the shadows = BD − BC $=15\sqrt{3}-5\sqrt{3}=10\sqrt{3}\mathrm{m}$

If the elevation of the sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high is $\underline{10\sqrt{3}\mathrm{m}}$.