Question

In the given figure, AB = AC. OB and OC bisect $\angle \text{ABC}$ and $\angle \text{ACB}$ respectively. If $\angle \text{BAC}={40}^{\circ }$, then $\angle \text{BOC}=$ ___

• A. ${110}^{\circ }$
• B. ${35}^{\circ }$
• C. ${140}^{\circ }$
• D. ${155}^{\circ }$

Answer 1

The correct option is A

${110}^{\circ }$

(i) $\Delta \text{ABC}$ is isosceles ..... (given, AB = AC)

(ii) $\angle \text{ABC}=\angle \text{ACB}=\frac{{180}^{\circ }-{40}^{\circ }}{2}$ ..... (Sum of angles of a $\Delta$)

$=\frac{{140}^{\circ }}{2}$

$={70}^{\circ }$

(iii) $\angle \text{ABO}=\angle \text{OBC}={35}^{\circ }$ ..... (BO bisects $\angle \text{ABC}$)

(iv) $\angle \text{ACO}=\angle \text{OCB}={35}^{\circ }$ ..... (CO bisects $\angle \text{ACB}$)

(v) In $\Delta \text{BOC}$

$\angle \text{BOC}+{35}^{\circ }+{35}^{\circ }={180}^{\circ }$ ..... (Sum of angles of a $\Delta$)

$\angle \text{BOC}={180}^{\circ }-{70}^{\circ }$

$={110}^{\circ }$