Răspuns:
Explicație pas cu pas:
Din T.Cosinusului, ⇒ AB²=AC²+BC²-2·AC·BC·cosB=2²+(√3 -1)²-2·2·(√3 -1)·cos120°=2²+(√3 -1)²-2·2·(√3 -1)·cos(180°-60°)=2²+(√3 -1)²-2·2·(√3 -1)·(-cos60°)=2²+(√3 -1)²+2·2·(√3 -1)·(1/2)=4+(√3)²-2√3·1+1²+2√3-2=4+3-2√3+1+2√3-2=6, deci AB=√6.
Din T.Sinusurilor, ⇒ AC/sinB=AB/sinC, ⇒ 2/sinC=√6/sin120°, ⇒ 2·sin120°=√6·sinB.
sin120°=sin(180°-120°)=sin60°=√3/2. Atunci, obtinem
2·(√3/2)=√6·sinB, ⇒ √3=√6·sinB, ⇒sinB=√3/√6=1/√2=√2/2
Atunci ∡B=45°
[tex]\it m(\hat A) +m(\hat B) =180^o-120^o =60^o \Rightarrow m(\hat A) =60^o-m(\hat B)\\ \\ Not\breve{a}m\ m(\hat B)=x\ \Rightarrow m(\hat A)=60^o-x\\ \\ Teorema\ sinusurilor \Rightarrow \dfrac{2}{sinx} =\dfrac{\sqrt3-1}{sin(60^o-x)} \Rightarrow \dfrac{sin(60^o-x)}{sinx}=\dfrac{\sqrt3-1}{2} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{sin60^o cosx-sinx cos60^o}{sinx}=\dfrac{\sqrt3}{2}-\dfrac{1}{2} \Rightarrow \dfrac{\sqrt3}{2}ctgx -\dfrac{1}{2}=\dfrac{\sqrt3}{2}-\dfrac{1}{2} \Rightarrow[/tex]
[tex]\it \Rightarrow \dfrac{\sqrt3}{2}ctgx=\dfrac{\sqrt3}{2} \Rightarrow ctgx=1 \Rightarrow x=45^o[/tex]
