Question

角AOE=150°,角AOB:角BOC=2:1,角COD:角DOE=1:2求角BOD的大小

Answer 1

设：角BOC=α,角COD=β,则 角AOB=2α,角DOE=2β

角BOD=α+β

3α+3β=150°那么 3（α+β）=150°

即：α+β=50°

答曰：求角BOD=50°

玩魔兽也会做算术!