Question

From a point P, two tangents PA and PB are drawn to a circle C(0, r).

If OP = 2r, then find ∠APB. What type of triangle is APB?

Answer 1

Let ∠APB = 2θ

By symmetry,

∠APO = ∠BPO

= 1/2 ∠APB

∠APO = ∠BPO = θ

Also,

We know that radius is perpendicular

\(\therefore\) ∠OAP = 90°

In right angle triangle △APO

Now we need to tell what type of triangle is APB

In △APB

Since tangents from external point is equal 

PA = PB

Thus,

∠PAB = ∠PBA

Thus,

In △APB

∠PAB = ∠PBA = 60°

And, ∠APB = 60°

Since all angles are 60°

\(\therefore\) △APB is an equilateral triangle.

Answer 2

let ∠APO = θ

Sinθ = \(\frac{OA}{OP}=\frac{1}{2}\)

⇒ θ = 30°

⇒ ∠APB = 2θ = 60°

Also ∠PAB = ∠PBA = 60° (∵ PA = PB)

⇒ △ APB is equilateral

Comments

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