Question

A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) 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∘. The height of the tower is​

Answer 1

let us consider the height of the tower 'h'

we know that 'h'= AC tan 30 degree = BC tan 30 degree

We know that AC=BC [∵ radius of circle]

Hence, we can say that △ABC is an isosceles triangle with AC=BC.

So, ∠ABC=∠BAC ...(2)

But ∠ACB=60

∘

[given]

⇒∠ABC+∠BAC+∠ACB=180

∘

⇒∠ABC+∠ABC+60

∘

=180

∘

[from (2)]

⇒2×∠ABC=120

∘

∴∠ABC=∠BAC=60

∘

Hence △ABC is an equilateral triangle.

∴AB=BC=CA=a ...(3)

∴h=atan30

∘

[form equation (1)]

=

3

a

Answer 2

Step-by-step explanation:

Solution

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Correct option is C)

Let us consider the height of the tower as h

We know that, h=ACtan30

∘

=BCtan30

∘

...(1)

We know that AC=BC [∵ radius of circle]

Hence, we can say that △ABC is an isosceles triangle with AC=BC.

So, ∠ABC=∠BAC ...(2)

But ∠ACB=60

∘

[given]

⇒∠ABC+∠BAC+∠ACB=180

∘

⇒∠ABC+∠ABC+60

∘

=180

∘

[from (2)]

⇒2×∠ABC=120

∘

∴∠ABC=∠BAC=60

∘

Hence △ABC is an equilateral triangle.

∴AB=BC=CA=a ...(3)

∴h=atan30

∘

[form equation (1)]

=

3

a