In ∆ ABC, the internal bisectors of ∠B and ∠C

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In ∆ ABC, the internal bisectors of ∠B and ∠C meet at point O. If ∠A = 80°, then ∠BOC is equal to :

We draw a figure triangle whose the internal bisectors of ∠B and ∠C meet at point O ,

∠OBC =	1	∠ABC
	2

∠OCB =	1	∠ACB
	2

∴ ∠OBC + ∠OCB =	1	(∠ABC + ∠ACB)
	2

∠OBC + ∠OCB =	1	(180° - ∠BAC)
	2

∠OBC + ∠OCB =	1	(180° - 80°)
	2

∠OBC + ∠OCB =	100°	= 50°
	2

∴ In ∆ OBC,
∠BOC = 180° – (∠OBC + ∠OCB)
Hence , ∠BOC = 180° – 50° = 130°