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如图,△ABC中,∠ABC、∠ACB的平分线交于点O。 ⑴若∠A=70°,求∠BOC的度数。

如图,延长AO交于BC于点D,∵∠B和∠C的平分线交于点O∴∠ACB=2∠2,∠ABC=2∠1,∵∠ACB+∠ABC+∠BAC=180°,∴2∠1+2∠2+∠BAC=180°,∴∠1+∠2=(180°-∠BAC)÷2=(180°-70°)÷2=55度.∵∠BOD=∠1+∠BAO,

∵∠A=70°,∴∠ABC+∠ACB=180°-∠A=110°,∵BO、CO分别是△ABC的角∠ABC、∠ACB的平分线,∴∠OBC=12∠ABC,∠OCB=12∠ACB,∴∠OBC+∠OCB=12(∠ABC+∠ACB)=55°,∴∠BOC=180°-(∠OBC+∠OCB)=180°-55°=125°.故答案为:125.

①在△ABC中,由∠A=70°,得∠ABC+∠ACB=110°,∵BO和CO分别平分∠ABC和∠ACB,∴∠OBC+∠OCB=12(∠ABC+∠ACB)=55°,在△OBC中,∠BOC=180°-(∠OBC+∠OCB)=125°.②在△ABC中,由∠A=n°,得∠ABC+∠ACB=18

(1)∵∠A=70°∴∠ABC+∠ACB=110°∴1/2∠ABC+1/2∠ACB=55°∵∠OBC=1/2∠ABC ∠OCB=1/2∠ACB∴∠OBC+∠OCB=55°∴∠BOC=125°(2))∵∠A=n°∴∠ABC+∠ACB=(180-n)°∴1/2∠ABC+1/2∠ACB=[(180-n)/2]°∵∠OBC=1/2

因为角A=70度,所以角B+角C=180-70=110度,又三角形BOC的底角BCO和CBO分别是角B角C的一半,所以角BOC=180-1/2*110 =125度.

∵ 三角形内角和为180°∴ ∠A+∠ABC+∠ACB=180°∴ ∠ABC+∠ACB=180°-76°=104°∵ BO与CO分别为∠ABC与∠ACB的角分线∴ 1/2∠ABC+1/2∠ACB= 1/2(∠ABC+∠ACB)=∠OBC+∠OCB=52°∵ 在△BOC中∴ ∠BOC+∠OBC+∠OCB=

∵∠BOC=132°,∴∠OBC+∠OCB=48°,∵∠ABC与∠ACB的平分线相交于O点,∴∠ABC=2∠OBC,∠ACB=2∠OCB,∴∠ABC+∠ACB=2(∠OBC+∠OCB)=96°,∴∠A=180°-96°=84°.同理,∵∠BOC=a°,∴∠OBC+∠OCB=180°-α°.∵∠ABC与∠ACB的平分线相交于O点,∴∠ABC=2∠OBC,∠ACB=2∠OCB,∴∠ABC+∠ACB=2(∠OBC+∠OCB)=2(180-α)°=360°-2α°,∴∠A=180°-360°+2α°=2α°-180°.

如图,在ABC中,∠ABC、∠ACB的平分线交于点O.若∠BOC=140°,则∠A= ( ) A.70° B.80° C.90° ..

(1)∵∠A=50°,∴∠ABC+∠ACB=180°-∠A=180°-50°=130°,∵∠ABC,∠ACB的角平分线相交于点O,∴∠OBC=12∠ABC,∠OCB=12∠ACB,∴∠OBC+∠OCB=12(∠ABC+

(1)∵BI是∠ABC的平分线,∠ABC=70°,∴∠CBI=12∠ABC=35°,∵CI是∠ACB的平分线,∠ACB=50°,∴∠BCI=12∠ACB=25°,在△BCI中,∠BIC+∠BCI+∠CBI=180°,

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