Question

已知od平分角aob,射线oc在 角aod的内部角boc等于二角aoc,角aob等于114度,求角

已知od平分角aob,射线oc在 角aod的内部角boc等于二角aoc,角aob等于114度,求角cod的度数

Answer 1

解：设∠AOC=α 则∠COB=2α
又∵ ∠AOB=114º
∴ 2α + α= 114º α=38º
又∵OD是 ∠AOB 的角平分线
∴ ∠AOD=57º
∴ ∠COD=57º-α=57º-38α=19º

Answer 2

∵OD平分∠AOB
∴∠AOD=∠BOD=1/2∠AOB=1/2×114°=57°
∵∠BOC=2∠AOC
∠AOC+∠BOC=∠AOB=114°
∴∠AOC+2∠AOC=114°
∠AOC=38°
∴∠COD=∠AOD-∠AOC=57°-38°=19°

Answer 3

BOC=2AOC, BOC=2/3AOB
BOD=1/2AOB, DOC+BOD= BOC,
BOD=1/6 AOB=114/6=19

Answer 4

∵OD平分∠AOB
∴∠AOD=∠BOD=1/2∠AOB=1/2×114°=57°
∵∠BOC=2∠AOC
∠AOC+∠BOC=∠AOB=114°
∴∠AOC+2∠AOC=114°
∠AOC=38°
∴∠COD=∠AOD-∠AOC=57°-38°=19°

Answer 5

∵OD平分∠AOB
∴∠AOD=∠BOD=1/2∠AOB=1/2×114°=57°
∵∠BOC=2∠AOC
∠AOC+∠BOC=∠AOB=114°
∴∠AOC+2∠AOC=114°
∠AOC=38°
∴∠COD=∠AOD-∠AOC=57°-38°=19°