Question

如图,(1)如图1在△ABC中,OB,OC分别是∠ABC,∠ACB的平分线,若∠A=x°,则∠BOC的度数为多少?

Answer 1

1.∠BOC+1/2(∠ABC+∠ACB)=180,

∠BOC=180-1/2(∠ABC+∠ACB),

若∠A=x°,

(∠ABC+∠ACB)+∠A=180,

(∠ABC+∠ACB)=180-∠A=180-x°,

∠BOC=180-1/2(∠ABC+∠ACB)=180-1/2(180-x°)=90+(x°/2).

2.∠A+∠B+∠AEB+∠ADB=360,

∠AEB+∠ADB=360-(∠A+∠B)=360-(x°+y°),

而,1/2(∠AEB+∠ADB)+∠DCE+∠B=360,

∠DCE=360-1/2(∠AEB+∠ADB)-∠DCE=180-(x°+3y°)/2.