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

Answer: Option A

Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that ∠ CAD = ∠ BAD = $$\theta$$
From "Angle Bisector Theorem", we have
$$\frac{BD}{DC}=\frac{AB}{AC}$$
$$\Rightarrow \frac{1}{9}=\frac{15}{AC}$$     [∵ BD : DC = 1 : 9 and AB = 15(given)]
$$\Rightarrow AC=15\times 9\text{m}......\left(eq:1\right)$$
$$\text{From}\text{the}\text{right}\Delta ABC,$$
$$CB=\sqrt{A{C}^2-A{B}^2}$$     (∵ Pythagorean theorem)
$$=\sqrt{{\left(15\times 9\right)}^2-{15}^2}$$     [∵AC=15 × 9(eq : 1) and AB=15 m(given)]
$$\begin{array}{rl}& =\sqrt{{15}^2\times 9^2-{15}^2}\\ & =\sqrt{{15}^2\left(9^2-1\right)}\\ & =\sqrt{{15}^2\times 80}\\ & =\sqrt{{15}^2\times 16\times 5}\\ & =15\times 4\times \sqrt{5}\\ & =60\sqrt{5}\text{m}\end{array}$$