Question

如图,直线AB与CD相交于点O,OE⊥AB,OF⊥CD,垂足为O

如果∠AOC=4分之1∠EOF，求∠AOC的度数。
图形

Answer 1

设∠AOC = α.则∠EOF = 4α.
∵∠AOC + ∠COE = 90°，∠AOC + ∠AOF = 90°.
∴∠COE = ∠AOF
而∠COE + ∠AOF = ∠EOF - ∠AOC = 3α.
∴∠COE = 3α/2.
∴∠AOC + ∠COE = α + 3α/2 = 5α/2.
∴5α/2 = 90°，α = 36°，即∠AOC = 36°.

Answer 2

设∠AOC = α.则∠EOF = 4α.
∵∠AOC + ∠COE = 90°，∠AOC + ∠AOF = 90°.
∴∠COE = ∠AOF
而∠COE + ∠AOF = ∠EOF - ∠AOC = 3α.
∴∠COE = 3α/2.
∴∠AOC + ∠COE = α + 3α/2 = 5α/2.
∴5α/2 = 90°，α = 36°，即∠AOC = 36°