In figure, if 0 is the center of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then POQ is equal to


Given : OP is a radius and PR is a tangent in a circle with center O with RPQ = 50°


To find : POQ


Now, OP PR [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]


OPR = 90°


OPQ + RPQ = 90°


OPQ + 50° = 90°


OPQ = 40°


In POQ


OP = OQ [radii of same circle]


OQP = OPQ = 40° [ angles opposite to equal sides are equal] [1]


In OPQ By angle sum property of a triangle


OPQ + OPQ + POQ = 180°


40° + 40° + POQ = 180°


POQ = 100°

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