From a point P, two tangents PA and PB are drawn to a circle with center O. If OP =

diameter of the circle shows that ΔAPB is equilateral.

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#### Solution

OP = 2r

Tangents drawn from external point to the circle are equal in length

PA = PB

At point of contact, tangent is perpendicular to radius.

In ΔAOP, sin 𝜃 =`"opp.side"/"hypotenuse"=r/(2r)=1/2`

𝜃 = 30°

∠APB = 20 = 60°, as PA = PB ∠BAP = ∠ABP = x.

In ΔPAB, by angle sum property

∠APB + ∠BAP + ∠ABP = 180°

2x = 120° ⇒ x = 60°

In this triangle all angles are equal to 60°

∴ ΔAPB is equilateral.

Is there an error in this question or solution?

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