In the given figure ∠A = 80° and the bisectors to interior angles �

1.	In the given figure ∠A = 80° and the bisectors to interior angles ∠B and ∠C meet at ‘O’. Then ∠BOC =A) 40° B) 130° C) 65°D) 160°
Answer» Correct option is (B) 130° \(\angle A=80^\circ\) \(\because\) \(\angle A+\angle B+\angle C=180^\circ\) \(\Rightarrow\) \(\angle B+\angle C=180^\circ-\angle A\) \(=180^\circ-80^\circ=100^\circ\) Given that angle bisectors of \(\angle B\) and \(\angle C\) meet at O. \(\therefore\) \(\angle OBC=\frac{\angle B}2\) & \(\angle OCB=\frac{\angle C}2\) \(\therefore\) \(\angle OBC+\angle OCB=\frac{\angle B}2+\frac{\angle C}2\) \(=\frac{\angle B+\angle C}2=\frac{100^\circ}2=50^\circ\) \(\because\) \(\angle OBC+\angle OCB+\angle BOC=180^\circ\) (Sum of angles in a triangle is \(180^\circ)\) \(\Rightarrow\) \(\angle BOC=180^\circ-(\angle OBC+\angle OCB)\) \(=180^\circ-50^\circ=130^\circ\) Correct option is B) 130°

In the given figure ∠A = 80° and the bisectors to interior angles ∠B and ∠C meet at ‘O’. Then ∠BOC =A) 40° B) 130° C) 65°D) 160°

Answer»

Correct option is (B) 130°

\(\angle A=80^\circ\)

\(\because\) \(\angle A+\angle B+\angle C=180^\circ\)

\(\Rightarrow\) \(\angle B+\angle C=180^\circ-\angle A\)

\(=180^\circ-80^\circ=100^\circ\)

Given that angle bisectors of \(\angle B\) and \(\angle C\) meet at O.

\(\therefore\) \(\angle OBC=\frac{\angle B}2\) & \(\angle OCB=\frac{\angle C}2\)

\(\therefore\) \(\angle OBC+\angle OCB=\frac{\angle B}2+\frac{\angle C}2\)

\(=\frac{\angle B+\angle C}2=\frac{100^\circ}2=50^\circ\)

\(\because\) \(\angle OBC+\angle OCB+\angle BOC=180^\circ\)

(Sum of angles in a triangle is \(180^\circ)\)

\(\Rightarrow\) \(\angle BOC=180^\circ-(\angle OBC+\angle OCB)\)

\(=180^\circ-50^\circ=130^\circ\)

Correct option is B) 130°