Q14 of 69 Page 9

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

∠OBC = 180o - ∠B - (180o - ∠B)


∠OBC = 90o - ∠B


And,


∠OCB = 180o - ∠C - (180o - ∠C)


∠OCB = 90o - ∠C


In


∠BOC + ∠OCB + ∠OBC = 180o


∠BOC + 90o - ∠C + 90o - ∠B = 180o


∠BOC = (∠B + ∠C)


∠BOC = (180o - ∠A) [From ]


∠BOC = 90o - ∠A


∠BOC = 90o -

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