Answer :

Given : A circle with center O and PA and PB are two tangents to the circle at point A and C from an external point P such that ∠APC = 60° [i.e. angle of inclination between two tangents] .

To Find : AP and PC

In △OAP and △OCP

AO = OC [ radii of same circle]

OP = OP [ common ]

AP = PC [ tangents through an external point to a circle are equal]

△OAP ≅ △OPC [ By Side Side Criterion]

∠APO = ∠OPC [Corresponding parts of congruent triangles are equal] [1]

Now, ∠APC = 60° [Given]

∠APO + ∠OPC = 60°

∠APO + ∠APO = 60° [By 1]

2∠AP0 = 60°

∠APO = 30°

Now, OA ⏊ AP [ As tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OAP = 90°

So △AOP is a right-angled triangle

And we know that,

So,

[As OA is radius and equal to 3 cm]

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