Question

如图，直线AB,CD相交于点O,射线OE⊥AB于O，射线OF⊥CD于O，且角AOF=25度。求角B

如图，直线AB,CD相交于点O,射线OE⊥AB于O，射线OF⊥CD于O，且角AOF=25度。求角BOC与角EOD的度数

Answer 1

∵OE⊥AB于O，OF⊥CD于O
∴∠AOE=∠DOF=90°
∴∠AOF+∠EOF=∠EOF+∠DOE=90°
∴∠DOE=∠AOF=25°
2. ∴∠AOD=∠AOE+∠DOE=90°+25°=115°
∵AB,CD相交于点O
∴∠BOC=∠AOD=115°

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