a)
m∡AOB=m∡AOM×2=60×2=120°
m∡BOD=360 -( 90+120)=360-210=150°
stm ca : m∡COD= 2×m∡BOC
⇒ m∡COD+ m∡BOC=m∡BOD= 3×m∡BOC
⇒m∡BOC=150 /3= 50°
b)
am vazut la punctul a ca m∡BOD= 3×m∡BOC=150°
m∡DOM= 90+60=150°
⇒ m∡DOM = m∡BOD
c) prelungim [OD] si notam [ON]....⇒ m∡DON=180°
m∡DON- m∡AD= m∡AON= 180-90= 90°
m∡AON-m∡AOM=m∡MON= 90-60=30°
m∡MOB=60°
m∡MOB-∡MON=30°
⇒ ∡MON=m∡NOB=30°
