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# 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?

A. $$60\sqrt 5 \,{\text{m}}$$

B. $${\text{15}}\sqrt 5 \,{\text{m}}$$

C. $$15\sqrt 3 \,{\text{m}}$$

D. $${\text{60}}\sqrt 3 \,{\text{m}}$$

### Solution(By Examveda Team)

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)]
\eqalign{ & = \sqrt {{{15}^2} \times {9^2} - {{15}^2}} \cr & = \sqrt {{{15}^2}\left( {{9^2} - 1} \right)} \cr & = \sqrt {{{15}^2} \times 80} \cr & = \sqrt {{{15}^2} \times 16 \times 5} \cr & = 15 \times 4 \times \sqrt 5 \cr & = 60\sqrt 5 \,{\text{m}} \cr}

This Question Belongs to Arithmetic Ability >> Height And Distance

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