Let the point where the ship turns be C.
A, B and C form a triangle whose have length 230, 210 and 180 miles.
According to the cosine Law:
[tex]|AB|^{2}= |AC|^{2}+|CB|^{2}-2|AC|*|CB|*cos(ACB)[/tex]
[tex]210^{2}=230^{2}+180^{2}-2*230*180*cos(ACB)[/tex]
[tex]44,100=52,900+32,400-82,800cos(ACB)[/tex]
[tex]44,100=85,300-82,800cos(ACB)[/tex]
[tex]82,800cos(ACB)=85,300-44,100[/tex]
[tex]cos(ACB)= \frac{85,300-44,100}{82,800}=0.5[/tex]
Thus, m(ACB)=60°,
thus x=120°
Answer: x=120°
