Question

In a triangle ABC, O is the center of incircle PQR, $\angle$BAC = ${65}^{\circ }$, $\angle$BCA = ${75}^{\circ }$, find $\angle$ ROQ.

• A. ${80}^{\circ }$
• B. ${120}^{\circ }$
• C. ${140}^{\circ }$
• D. Can't be determined

Answer 1

The correct option is C

${140}^{\circ }$

Given, $\angle A={65}^{o}and\angle B=5^o$.

In $△ABC$,
$\angle A+\angle B+\angle C={180}^{\circ }$
(Angle sum property of a triangle)
${65}^{\circ }+\angle B+{75}^{\circ }={180}^{\circ }$
$\angle B={40}^{\circ }$

We know, $\angle ORB={90}^{\circ }$ and
$\angle OQB={90}^{\circ }$ .
(Line joining point of contact of tangent to centre of circle is perpendicular to tangent.)

In quadrilateral ROQB,
$\angle ROQ+\angle OQB+\angle QBR+\angle BRO={360}^o$
[Angle sum property of a quadrilateral]
$\angle ROQ+{90}^o+{40}^o+{90}^o={360}^o$
$\angle ROQ={360}^o-{220}^o$
$={140}^{\circ }$