Question

$\text{If AB is the tangent to the circle with}\text{center O then, find the measure of}\angle OCP.$
$\text{Given that OP = PC.}$

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is B

${45}^{\circ }$

The tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\therefore \angle OPC={90}^{\circ }$

$\text{Given, OP = PC.}$

$\text{So,}△OPC\text{is an isosceles right}\text{angled triangle.}\Rightarrow \angle PCO=\angle POC$

$\angle PCO+\angle POC+\angle OPC={180}^{\circ }\text{(Angle sum property of a triangle)}$

$\angle PCO+\angle POC+{90}^{\circ }={180}^{\circ }$

$\angle PCO+\angle POC={90}^{\circ }$

$\text{Hence,}\angle POC=\angle OCP={45}^{\circ }$