Răspuns:
4) < AOB + < BOC+ < COD+< DOA= 360°
m ( < BOC) = 2× m ( < AOB)
m ( < COD) = 3 ×m ( < AOB)
m ( < DOA) = 6× m (< AOB)
< AOB + 2m (<AOB)+ 3m (< AOB )+6 m(< AOB)= 360 °
12 m ( < AOB) = 360°
AOB= 360°÷12
m < AOB= 30°
m < BOC= 2×30°=60°
m < COD= 3 × 30°= 90°
m < DOA= 6× 30°= 180°
5)
m ( < BOC) = 90 °
m( < DOF) = 32° 》 m < BOF = BOD-m (< DOF) - sunt unghiuri complementare
90°-32° = 58°
m<AOC= 90°
