Trigonometry! #74

Geometry Level 3

A tower stands at the centre of a circular park. A A and B B are two points on the boundary of the park such that A B = a AB = a subtends an angle of 6 0 ∘ 60^{\circ} at the foot of the tower and the angle of elevation of the top of the tower from A A or B B is 3 0 ∘ 30^{\circ} .

If the height of the tower, in terms of a a , is x x , then enter the value of x a \frac {x}{a} correct to three decimal places.

This problem is part of the set Trigonometry .


The answer is 0.577.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Let the foot of the tower be O O and its top, P P . Since ∠ A O B = 6 0 ∘ \angle AOB = 60^\circ , △ A O B \triangle AOB is an equilateral triangle implying O A = O B = A B = a OA=OB=AB=a . It is given that ∠ O A P = 3 0 ∘ \angle OAP = 30^\circ implying that O P O A = x a = tan ⁡ 3 0 ∘ = 1 3 = 0.577 \frac {OP}{OA} = \frac {x}{a} = \tan {30^\circ} = \frac {1}{\sqrt{3}} = \boxed{0.577} .

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Noel Lo
Apr 28, 2015

Same as Chew-Seong's method.

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