Question

点o是三角形abc的外心,角a＝72度,求角boc的度数

Answer 1

点o是三角形abc的外心,则oa=ob=oc,∠oab=∠oba,∠oac=∠oca,∠oab+∠oac=∠a＝72度,∠boc=∠oab+∠oba+∠oac+∠oca=144度.