Given,
ABC is a triangle
∠A = 72° and internal bisectors of B and C meet O.
In ΔABC
∠A + ∠B + ∠C = 180°
72° + ∠B + ∠C = 180°
∠B + ∠C = 180° – 72°
∠B + ∠C = 108°
Divide both sides by 2, we get
\(\frac{∠B}{2}\) + \(\frac{∠C}{2}\) = \(\frac{108}{2}\)
\(\frac{∠B}{2}\) + \(\frac{∠C}{2}\) = 54°
∠OBC + ∠OCB = 54° (i)
Now, in ΔBOC
∠OBC + ∠OCB + ∠BOC = 180°
54° + ∠BOC = 180°[Using (i)]
∠BOC = 180° – 54°
= 126°
