In the given figure, AB is a tangent to the circle with centre O. If OP = PC, then find $∠ O C P$ .

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Solution

Since a tangent at any point of a circle is perpendicular to the radius at the point of contact, we have $∠$ OPC $= 90^{\circ}$ . [Using Tangent Theorem] [0.5 Marks]

Since OP = PC, $△$ OPC is an isosceles right-angled triangle. Hence, $∠$ POC $= ∠$ OCP $= 45^{\circ}$ . [0.5 Marks]

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