Question

$\text{PA and PB are the two tangents drawn}\text{to the circle. O is the center of the circle.}\text{A and B are the points of contact of the}\text{tangents PA and PB with the circle.}\text{If}\angle OPA={35}^{\circ },\text{then find}\angle POB.$

• A. ${55}^{\circ }$
• B. ${65}^{\circ }$
• C. ${75}^{\circ }$
• D. ${85}^{\circ }$

Answer 1

The correct option is A

${55}^{\circ }$

$\text{Given, PA and PB are two tangents.}\text{And}\angle OPA={35}^o.$

We know that line joining point of contact of tangent to centre of circle is perpendicular to tangent.
$\therefore \angle OAP=\angle OBP={90}^{\circ }...\left(i\right)$

$\text{In,}△OAP,\text{by angle sum property of}\text{a triangle,}$
$\angle OAP+\angle APO+\angle POA={180}^o$
$\angle AOP={180}^{\circ }-{35}^{\circ }-{90}^{\circ }$
$\angle AOP={55}^{\circ }$

$\text{Consider}△OAP\text{and}△OBP$,
$OA=OB\left(\text{radius}\right)$
$\angle OAP=\angle OBP={90}^{\circ }\text{(from (i))}$
$PA=PB\text{(Tangents from an external}\text{point to the circle are equal in length.)}$

$\therefore △OAP\cong △OBP\text{(By SAS congruence criterion)}$

$\text{By}CPCT,\angle AOP=\angle BOP.$
$\text{Thus,}\angle BOP={55}^{\circ }$.