Answer :

Let AP and BP are two tangents from an external point P on a circle with center O, at points A and B respectively.

Angle with AP and BP,

∠APB = 60° [Given]

In ΔAOP and ΔBOP

AP = BP [Tangents drawn from an external point to a circle are equal]

OP = OP [Common]

OA = OB [Radii of same circle]

ΔAOP ≅ ΔBOP [By Side-Side-Side Criterion]

∠OPA = ∠OPB [Corresponding parts of congruent triangles are equal]

Also,

∠OPA + ∠OPB = ∠APB

⇒ ∠OPA + ∠OPA = 60°

⇒ 2∠OPA = 60°

⇒ ∠OPA = 30°

Also, OA ā AP [Tangent drawn at a point on the circle is perpendicular to the radius through point of contact]

In ΔOAP,

[As OA = radius of circle = a]

⇒ OP = 2a

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