PQ is a tangent to a circle with center O at the point P. If ΔOPQ is an isosceles triangle, then ∠OQP is equal to
Let us consider a circle with center O and PQ is a tangent
on the circle, Joined OP and OQ
But OPQ is an isosceles triangle, ∴ OP = PQ
∠OQP = ∠POQ
[Angles opposite to equal sides are equal]
In △OQP
∠OQP + ∠OPQ + ∠POQ = 180°
[Angle sum property of triangle]
∠OQP + 90° + ∠OPQ = 180°
2 ∠OPQ = 90°
∠OPQ = 45°