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The top of a 15 metre high tower makes an angle of elevation of \( \Large 60 ^{\circ} \) with the bottom of an electric pole and an angle of elevation of \( \Large 30 ^{\circ} \) with the top of the pole. What is the height of the electric pole ?
A) 5 metres
B) 8 metres
C) 10 metres
D) 12 metres
Correct answer:
C) 10 metres
Description for Correct answer:
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
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