The angle of elevation of the top of a tower at a point on the ground is \( \Large 30 ^{\circ} \). If on walking 20m towards the tower the angle of elevation becomes \( \Large 60 ^{\circ} \) then the heights of the tower is:

A) 10 m

B) \( \Large \frac{10}{\sqrt{3}} m \)

C) \( \Large 10\sqrt{3} m \)

D) None of these

Correct answer:
C) \( \Large 10\sqrt{3} m \)

Description for Correct answer:

Let the height of the tower be h.



\( \Large \triangle PAO, \tan 60 ^{\circ} = \frac{h}{OA} \)

=> \( \Large OA = h \cot 60 ^{\circ} \)

= \( \Large \frac{h}{\sqrt{3}} \)

\( \Large In \triangle PBO, \tan 30 ^{\circ} = \frac{h}{OB} \)

=> \( \Large OB = \frac{h}{\frac{1}{\sqrt{3}}} = \sqrt{3}h \)

=> \( \Large AB + AO = \sqrt{3}h \)

=> \( \Large 20 + \frac{h}{\sqrt{3}} = \sqrt{3}h \)

=> \( \Large h = \frac{20}{\sqrt{3}-\frac{1}{\sqrt{3}}} = \frac{20\sqrt{3}}{2} = 10\sqrt{3} m \)



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