Question

In the given figure,
BAC = 50°, the bisectors of exterior angles B and C meet 'O'. Find angle BOC.And also prove the theorem 90°- angle A​

Answer 1

∠BOC = 90° - ∠A/2 , ∠BOC = 65°   if ∠BAC = 50°

Step-by-step explanation:

exterior ∠B = ∠A + ∠C

=> (1/2) exterior ∠B =  (1/2) (∠A + ∠C)

exterior ∠C = ∠A + ∠B

=> (1/2) exterior ∠C =  (1/2) (∠A + ∠B)

Bisector of exterior ∠B + Bisector of exterior ∠C  + ∠BOC = 180°

=>  (1/2) (∠A + ∠C) + (1/2) (∠A + ∠B)  + ∠BOC = 180°

=> ∠A + (1/2)(∠B + ∠C) + ∠BOC = 180°

∠A + ∠B + ∠C = 180°  => ∠B + ∠C = 180° - ∠A

=> ∠A +  (1/2)(180° - ∠A) + ∠BOC = 180°

=> ∠BOC = 90° - ∠A/2

∠BOC = 90° - 50°/2

=> ∠BOC = 65°

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