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