If AB is the tangent to the circle with center O what is $∠ O C P$ ? Given that OP= PC

$30^{\circ}$

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$45^{\circ}$

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$60^{\circ}$

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$15^{\circ}$

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Solution

The correct option is B
$45^{\circ}$

If AB is tangent then $∠$ OPC = $90^{\circ}$ (point of contact of radius and tangent)
And given that OP = PC,

It makes $Δ$ OPC an isosceles right triangle hence $∠$ POC = $∠$ OCP = $45^{\circ}$

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