A body weighing $13\text{kg}$ is suspended by two strings $5\text{m}$ and $12\text{m}$ long, their other ends being fastened to the extremities of a rod $13\text{m}$ long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are:

Let the rod be $AB$ of length $13\text{m}$. Let the body be at $O$.The strings are $OA=5\text{m}$ and $OB=12\text{m}$.In $△OAB$, $O{A}^2+O{B}^2=5^2+12^2=25+144=169=13^2=A{B}^2$.Thus, $△OAB$ is a right-angled triangle at $O$.Since the body hangs immediately below the midpoint $C$ of the rod, $OC$ is vertical.The angle $\theta$ between the tension $T_1$ (along $OA$) and the vertical $OC$ is such that:In $△OAC$, since $C$ is the midpoint of the hypotenuse, $OC=CA=CB=6.5\text{m}$.The triangle $OAC$ is isosceles, so $\angle COA=\angle CAO=A$.Similarly, $\angle COB=\angle CBO=B$.Applying Lami's Theorem or resolving forces:$T_1\sin A=T_2\sin B$ is not needed here; rather $T_1=W\cos B$ and $T_2=W\cos A$ for a right-angled geometry like this.$\cos A=\frac{5}{13}$, $\sin A=\frac{12}{13}$.$\cos B=\frac{12}{13}$, $\sin B=\frac{5}{13}$.The tensions are $T_1=13\times \frac{5}{13}=5\text{kg}$ and $T_2=13\times \frac{12}{13}=12\text{kg}$.