Question

The top of a $15$ metre high tower makes an angle of elevation of ${60}^0$ with the bottom of an electric pole and angle of elevation of ${30}^0$ 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

Answer 1

The correct option is C $10$ metres

Let $AB$ be the tower and $CD$ be the electric pole
Then $\angle ACB={60}^0,\angle EDB={30}^0$ and $AB=15$ m
Let $CD=h.$ Then $BE=(AB-AE)$
$=(AB-CD)=(15-h)$
We have $\frac{AB}{AC}=\tan {60}^0=\sqrt{3}$
$\Rightarrow AC=\frac{AB}{\sqrt{3}}=\frac{15}{\sqrt{3}}$
And $\frac{BE}{DE}=\tan {30}^0=\frac{1}{\sqrt{3}}$
$\Rightarrow DE=(BE\times \sqrt{3})$ $=\sqrt{3}(15-h)$
$\Rightarrow 3h=(45-15)$

$\Rightarrow h=10$ m