In order to find the tension in the wire, let's first decompose it in its vertical and horizontal components:
[tex]\begin{gathered} T_x=T\cdot\cos (35\degree) \\ T_y=T\cdot\sin (35\degree) \end{gathered}[/tex]
Now, since the system is stable, the sum of vertical forces is equal to zero, so we have:
[tex]\begin{gathered} T_y-175-215=0 \\ T_y-390=0 \\ T_y=390 \\ T\cdot\sin (35\degree)=390 \\ T\cdot0.57358=390 \\ T=\frac{390}{0.57358} \\ T=679.94\text{ N} \end{gathered}[/tex]
So the tension in the wire is equal to 679.94 N.
