Question

The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

Answer 1

Let the height of the electric pole AD be “h” m

EC = 15 – h m

Let AB be “x”

In the right ∆ABC, tan 60° = ${\displaystyle \frac{\text{BC}}{\text{AB}}}$

${\displaystyle \sqrt{3}=\frac{15}{x}}$

x = ${\displaystyle \frac{15}{\sqrt{3}}}$

= ${\displaystyle \frac{15\times \sqrt{3}}{3}}$

= ${\displaystyle 5\sqrt{3}}$

In the right ∆CDE, tan 30° = ${\displaystyle \frac{\text{EC}}{\text{DE}}}$

${\displaystyle \frac{1}{\sqrt{3}}=\frac{15-\text{h}}{x}}$  ...(1)

Substitute the value of x = ${\displaystyle 5\sqrt{3}}$ in (1)

${\displaystyle \frac{1}{\sqrt{3}}=\frac{15-\text{h}}{5\sqrt{3}}}$

⇒ ${\displaystyle \sqrt{3}(15-\text{h})=5\sqrt{3}}$

(15 – h) = ${\displaystyle \frac{5\sqrt{3}}{\sqrt{3}}}$

⇒ 15 – h = 5

h = 15 – 5 = 10
∴ Height of the electric pole = 10 m