a)Daca,m(∡COB)=60°=>ΔCOB-echilat.=>CO=OB=BC=4√3
=>Diag.AC=2x4√3=8√3
In,ΔABC,dr,aflam,AB,cu,t.Pit.
AB²=AC²-BC²
AB²=192-48
AB²=144
AB=12cm
b)Se,formeaza,trapezul,dreptunghic,NDAM
Coboram,NE_l_AM(E∈AM),DN=AE=2cm,AD=NE=4√3cm(fiind,drepte,ll)
=>EM=4cm
Cu,t.Pit.,in,Δdr.NEM,aflam,NM
NM²=NE²+EM²
NM²=48+16
NM²=64
NM=8cm