Question

如图5-4,直线AB、CD、EF相交于同一点O,而且∠BOC=2/3∠AOC,∠DOF=1/3∠AOD,那么∠FOC=（

Answer 1

∵BOC=2/3∠AOC

∠BOC+∠AOC=180°

∠AOC=108°

∠BOC=72°

∵∠AOD=∠BOC=72°

∠DOF=1/3∠AOD=1/3（∠AOD+∠DOF）

∴∠DOF=1/3×72°+1/3∠DOF

∠DOF=36°

∴∠FOC=180°-∠DOF=180°-36°=144°