Correct Answer: 10 metres

Now, according to question,

AB = Height of Tower

= 15 metres

Let, CD = Height of Electric Pole = h metres

\( \Large \angle ADE = 30^{\circ} \)

\( \Large \angle ACB = 60^{\circ} \)

AE = (15 - h) metres

Distance between B and C = x metres

We know that

\( \Large \tan \theta = \frac{Perpendicular}{Base} \)

\( \Large \tan 30^{\circ} = \frac{15 - h}{x} \)

=> \( \Large \frac{1}{\sqrt{3}} = \frac{15 - h}{x} \)

\( \Large \therefore x = \sqrt{3} (15 - h) \) .... (i)

Now, \( \Large \tan 60^{\circ} = \frac{15}{x} => \sqrt{3} = \frac{15}{x} \)

\( \Large \therefore x = \frac{15}{\sqrt{3}} \) .... (ii)

From equation (i) and (ii)

\( \Large \sqrt{3}(15 - h) = \frac{15}{\sqrt{3}} \)

=> \( \Large 15 - h = \frac{15}{3} \)

=> h = 15 - 5 = 10 metres