Question

[Topic: Height and Distances]

A tower stands at the centre of a circular park. If A and B are two points on the boundary of the park, such that AB = a m subtends an angle of 60° at the foot of the tower and the angle of elevation of the top of the tower from A or B is 30°. Find the height of the tower.​

Answer 1

Answer:

a/√3

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Answer 2

Given :-

A tower stands at the centre of a circular park. If A and B are two points on the boundary of the park, such that AB = a m subtends an angle of 60° at the foot of the tower and the angle of elevation of the top of the tower from A or B is 30°.

To Find :-

Height of tower

Solution :-

Let height of tower = h

We know that radius of a circle is always equal.

AC = BC.

Then

∠ABC = ∠BAC

Since two sides are equal it must be an isosceles triangle.

Now,

In ΔABC

⇒ ∠ABC + ∠BAC + ∠CAB = 180

⇒ ∠ABC + ∠ABC + 60 = 180

⇒ 2∠ABC = 180 - 60

⇒ 2∠ABC = 120

⇒ ∠ABC = 120/2

⇒ ∠ABC = 60°

⇒ ∠ABC = ∠BAC = ∠BCA = 60°

It must be an equilateral triangle.

We know that sides of equilateral triangle are equal

AB = BC = BA

Let the side of equilateral triangle be x

Now

h = AC tan (30)° = BC tan (30)°

⇒ h = BC tan(30)°

⇒ h = x × 1/√3

⇒ h = x/√3

Hence,

The height of tower is x/√3 m