m∡(A)+m∡(B)+m∡(C)=180°⇒m∡(C)=180°-(80°+60°)=40°
deoarece BM=bisectoare⇒m∡(ABM)=m∡(MBC)=1/2m∡(B)=1/2·60°=30°
in ΔBMC avem:m∡(MBC)+m∡(BCM)+m∡(CMB)=180°⇒m∡(CMB)=180°-(40°+30°)=180°-70°=110°
in ΔABM avem:m∡(ABM)+m∡(BAM)+m∡(AMB)=180°⇒∡(BMA)=180°-(80°+30°)=180°-110°=70°