A body of mass 5kg is suspended by the strings making angles 60⁰ and 30⁰ with the horizontal as shown in figure (g =10m / s²)

The FBD of the block is as shown in fig

For equilibrium along X-axis
$\begin{array}{l}{\mathrm{T}}_2\cos 60={\mathrm{T}}_1\cos 30\\ \frac{{\mathrm{T}}_2}{2}=\frac{{\mathrm{T}}_1\sqrt{3}}{2}\Rightarrow {\mathrm{T}}_2=\sqrt{3}{\mathrm{T}}_1\end{array}$
For equilibrium along Y-axis,
$\begin{array}{l}{\mathrm{T}}_2\sin 60={\mathrm{T}}_1\sin 30=5\mathrm{g}\\ \frac{{\mathrm{T}}_2\sqrt{3}}{2}+\frac{{\mathrm{T}}_1}{2}=50\Rightarrow {\mathrm{T}}_2\sqrt{3}+{\mathrm{T}}_1=100\end{array}$
Solving above equations, we get
${\mathrm{T}}_1=25\mathrm{N};{\mathrm{T}}_2=25\sqrt{3\mathrm{N}}$