Question

如图,已知直线ab与cd相交于点o,oe⊥cd,of⊥ab,∠dof=3∠bod,求∠aoe的度数?

Answer 1

因为OF⊥AB，则∠FOB=90°，又因∠DOF=3∠DOB，所以∠BOD=30°=∠AOC
因为OE⊥CD，则∠COE=90°，所以∠AOE=∠COE+∠AOC=120°

Answer 2

∠DOF=3∠DOB
∠FOB = 90° = ∠DOF+∠DOB = 4∠DOB
所以∠DOB = 90/4 = 22.5 = ∠AOC
∠AOE = ∠AOC + ∠COE = 22.5+90 = 112.5

Answer 3

90*4+90=112.5