Question

In ∆ABC, the angle bisectors of ∠B and ∠C meet at O. If ∠A = 70°, find ∠BOC.

Answer 1

Let
$$\begin{gathered} \angle ABO=\angle OBC=\angle 1\left[\because OB\text{is the angle bisector}\right]\text{and} \\ \angle ACO=\angle OCB=\angle 2\left[\because OC\text{is the angle bisector}\right] \end{gathered}$$

From $∆ABC$, we have:
$\angle A+\angle B+\angle C=180°\left[\text{Sum of the angles of a triangle}\right]$
$$\begin{gathered} \Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle B+\frac{1}{2}\angle C=90° \\ \Rightarrow \frac{1}{2}\angle A+\angle 1+\angle 2=90°\left[\text{Here,}\frac{1}{2}\angle B=\angle ABO\text{and}\frac{1}{2}\angle C=\angle ACO\right] \\ \Rightarrow 35°+\angle 1+\angle 2=90° \\ \Rightarrow \angle 1+\angle 2=55°...\left(i\right) \end{gathered}$$

Now in $∆OBC$, we have:
$$\begin{gathered} \angle 1+\angle 2+\angle BOC=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow \angle 1+\angle 2+\angle BOC=180° \\ \Rightarrow 55°+\angle BOC=180°\left[\text{Using}\left(i\right)\right] \\ \Rightarrow \angle BOC=125° \end{gathered}$$