更新时间2019-09-17 09:20:34
,∠A=90°,求∠BOC
解:
设∠EBC为∠1,∠BCO为∠2,则:
已知∠A=90°,
∴∠ABC+∠ACB=90°,
∴∠ACB=90°-∠ABC
又∵CO是∠ACB的平分线,
∴∠2=½(90°-∠ABC)=45°-½∠ABC
∠DBC=180°-∠ABC,且BE是∠DBC的平分线,
∴∠1=½∠DBC=½(180°-∠ABC)=90°-½∠ABC
又∵∠1是∠OBC的外角,
∴∠1=∠BOC+∠2
∴∠BOC=∠1-∠2
=(90°-½∠ABC)-(45°-½∠ABC)
=90°-½∠ABC-45°+½∠ABC
=90°-45°
=45°
答:∠BOC为45°。
解:∠BOC=180°-∠ACB/2-∠CBO
=180°-∠ACB/2-∠CBA-∠ABO
=180°-∠ACB/2-∠CBA-∠DBE
=180°-∠ACB/2-∠CBA-(180°-∠CBA)/2
=180°-∠ACB/2-∠CBA-90°+∠CBA/2
=90°-(∠ACB+∠CBA/)2
=90°-90°/2
=45°
∠BOC=∠CBE-∠OCB
∠CBD=∠ACB+90
∠CBE=∠BCO+45
∠BOC=45