Free body diagram:
Here, T_1 is the tension in the left cable (T_1=30000 N; given), T_2 is the tension in the right cable, W is the weight of the tram.
The force equation in horizontal direction is given as,
[tex]T_1=T_2\sin (75\degree)[/tex]
Therefore, the tension in right cable T_2 is given as,
[tex]T_2=\frac{T_1}{\sin(75\degree)}[/tex]
Substituting all known values,
[tex]\begin{gathered} T_2=\frac{30000\text{ N}}{\sin (75\degree)} \\ \approx31058.28\text{ N} \end{gathered}[/tex]
The force equation in the vertical direction is given as,
[tex]W=T_2\cos (75\degree)[/tex]
Substituting all known values,
[tex]\begin{gathered} W=(31058.28\text{ N})\times\cos (75\degree) \\ \approx8038.47\text{ N} \end{gathered}[/tex]
Therefore, the weight of the tram is 8038.47 N.
