In a Δ ABC, the sides AB and AC are produced to P and Q

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In a Δ ABC, the sides AB and AC are produced to P and Q respectively. The bisectors of ∠ OBC and ∠ QCB intersect at a point O. Then ∠ BOC is equal to:

From given figure , We have ∠B + ∠CBP = 180° (Linear pair)

⇒	1	∠B + ∠CBP = 90°
	2

⇒	1	∠B + ∠1 = 90°
	2

⇒ ∠1 = 90° -	1	∠B ............ ( 1 )
	2

Similarly ∠2 = 90° -	1	∠C
	2

In ΔOBC, we have ∠1 + ∠2 + ∠BOC = 180° (Angle sum prop.)

⇒

90° -

∠B

90° -

∠C

+ ∠BOC = 180°

⇒ ∠BOC =	1	(∠B + ∠C)
	2

∠BOC =	1	(∠A + ∠B + ∠C) -	1	∠A
	2		2

∠BOC =	1	×180° -	1	∠A { ∴ ∠A + ∠B + ∠C = 180° }
	2		2

∠BOC = 90° -	1	∠A
	2

120° +	1	∠A
	2

120° -	1	∠A
	2