Answer :

**Given:** PA and PB are tangent of a circle PA= 10 cm and angle APB= 60^{o}

Let O be the center of the given circle and C be the point of intersection of OP and AB

In triangle PAC and triangle PBC

PA = PB (tangent from an external point are equal)

APC =BPC (tangent from an external point are equally inclined to the segment joining center to the point)

PC =PC (common)

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