The bisectors of exterior angles at B and C of Δ ABC meet at O, if ∠A= x°, then ∠BOC =

∠OBC = 180^o - ∠B - (180^o - ∠B)

∠OBC = 90^o - ∠B

And,

∠OCB = 180^o - ∠C - (180^o - ∠C)

∠OCB = 90^o - ∠C

In

∠BOC + ∠OCB + ∠OBC = 180^o

∠BOC + 90^o - ∠C + 90^o - ∠B = 180^o

∠BOC = (∠B + ∠C)

∠BOC = (180^o - ∠A) [From ]

∠BOC = 90^o - ∠A

∠BOC = 90^o -