Question

如图,已知三角形ABC中,三条内角平分线AD,BE,CF相交于点O,OG垂直于BC,垂足为G.若角ABC=32°,角AC

Answer 1

∵∠ABC＝32,∠ACB＝76

∴∠BAC＝180-∠ABC-∠ACB＝72

∵AD平分∠BAC

∴∠CAD＝∠BAC/2＝36

∵CF平分∠ACB

∴∠ACF＝∠BCF＝∠ACB/2＝38

∴∠COD＝∠CAD+∠ACF＝36+38＝74

∵OG⊥BC

∴∠COG+∠BCF＝90

∴∠COG＝90-∠BCF＝90-38＝52

∴∠DOG＝∠COD-∠COG＝74-52＝22°