Given, ∠A = 70°

Let the two angles ∠B = 2x and ∠C = 2y.

Then, angle bisector of B, ∠OBC = x and angle bisector of C, ∠OCB = y

⸫ ∠A + ∠B + ∠C = 180° [Sum of all angles of a triangle = 180°]

⇒ 70° + 2x + 2y = 180°

⇒ 2x + 2y = 110°

⸫ x + y = 55° …. (i)

Now,

∠BOC + x + y = 180° [Sum of all angles of a triangle = 180°]

⇒ ∠BOC = 180° - (x + y)

⇒ ∠BOC = 180° - 55° [from eq. (i)]

⸫ ∠BOC = 125°