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°
• B. 120°
• C. 140°
• D. Can't be determined

Answer 1

The correct option is C

140°

In triangle ABC

$\angle$A + $\angle$B + $\angle$C = ${180}^{\circ }$

${65}^{\circ }$ + $\angle$B + ${75}^{\circ }$ = ${180}^{\circ }$

$\angle$B = ${40}^{\circ }$

We know that $\angle$ORB = ${90}^{\circ }$ and $\angle$OQB = ${90}^{\circ }$

$\angle$ROQ = ${180}^{\circ }$ - $\angle$B

$\angle$ROQ = ${180}^{\circ }$ – ${40}^{\circ }$ = ${140}^{\circ }$