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

$30^{\circ}$

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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
$45^{\circ}$

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

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

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

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

$If AB is the tangent to the circle with center O then, find the measure of ∠ O C P .$
$Given that OP = PC.$

In the given figure, O is the centre of a circle and $∠ A O C = 120^{\circ} .$ Then, $∠$ BDC = ?

(a) $60^{\circ}$

(b) $45^{\circ}$

(d) $15^{\circ}$

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