Physics - NLM Question with Solution | TestHub

Free body diagram of the rope: tensions $T_A$ at angle ${\theta }_1$ and $T_B$ at angle ${\theta }_2$ to the horizontal, weight $Mg$.

Horizontal: $T_A\cos {\theta }_1=T_B\cos {\theta }_2$. Vertical: $T_A\sin {\theta }_1+T_B\sin {\theta }_2=Mg$.

Eliminating $T_B$: $T_A=\frac{Mg\cos {\theta }_2}{\sin ({\theta }_1+{\theta }_2)}$. The tension is maximum at A, so the maximum tension is $\frac{Mg\cos {\theta }_2}{\sin ({\theta }_1+{\theta }_2)}$.