如图,已知直线AB,CD,EF相交于点O,EF⊥AB,OG为角COF的平分线,OH为角DOG的平分线,若角AOC:角DO
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∵EF⊥AB

∴∠AOF=90

∴∠COF=∠AOF-∠AOC=90-∠AOC

∵OG平分∠COF

∴∠COG=1/2∠COF=1/2(90-∠AOC)=45-1/2∠AOC

∴∠DOG=180-∠COG=180-(45-1/2∠AOC)=135+1/2∠AOC

∵OH平分∠DOG

∴∠DOH=1/2∠DOG=1/2(135+1/2∠AOC)=135/2+1/4∠AOC

∵∠AOC:∠DOH=8:29

∴∠DOH=29/8∠AOC

∴135/2+1/4∠AOC=29/8∠AOC

∴∠AOC=20

∴∠DOH=29/8∠AOC=145/2=72.5

∴∠COH=180-∠DOH=180-72.5=107.5