Question

A vertical pole fixed to the ground is divided in the ratio 1 $:$ 9 by a mark on it with the lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, $15m$ away from the base of the pole, what is the height of the pole?

• A. $60\sqrt{5}m$
• B. $15\sqrt{5}m$
• C. $15\sqrt{3}m$
• D. $60\sqrt{3}m$

Answer 1

The correct option is A $60\sqrt{5}m$


Let CB be the pole and point D divide it such that BD $:$ DC = 1 $:$ 9.
Given that $AB=15m$
The two parts subtend equal angles at point A such that :- $\angle CAD=\angle BAD=\theta$

From "Angle Bisector Theorem", we have,
$\frac{BD}{DC}=\frac{AB}{AC}$
$\Rightarrow \frac{1}{9}=\frac{15}{AC}$ [ $\because$ BD $:$ DC = 1 $:$ 9 and AB = 15]
$\Rightarrow AC=15\times 9$ $\dots \left(1\right)$

From the right $\Delta ABC$,
$CB=\sqrt{{AC}^2-{AB}^2}$
$\Rightarrow CB=\sqrt{{(15\times 9)}^2-{15}^2}$
$\Rightarrow CB=\sqrt{{15}^2\times 9^2-{15}^2}$
$\Rightarrow CB=\sqrt{{15}^2(9^2-1^2)}$
$\Rightarrow CB=\sqrt{{15}^2\times 80}$
$\Rightarrow CB=\sqrt{{15}^2\times 16\times 5}$
$\Rightarrow CB=15\times 4\times \sqrt{5}=60\sqrt{5}m$