Answer
$\vec {T_1} =-56.5i+67.4j$ and $\vec {T_2} =56.5i+32.6j$.
Work Step by Step
Step 1. Identify the vector of the weight as $\vec {F_w} =-100j$
Step 2. Identify the vector of T1 as $\vec {T_1} =-T_1cos50^{\circ} i+T_1sin50^{\circ} j\approx-0.643T_1i+0.766T_1j$
Step 3. Identify the vector of T2 as $\vec {T_2} =T_2cos30^{\circ} i+T_2sin30^{\circ} j\approx0.866T_2i+0.5T_2j$
Step 4. For the forces to be in equilibrium, the total force should be zero $\vec {F} + \vec {T_1} + \vec {T_2}=0$
which gives $0.866T_2=0.643T_1$ (or $T_2=0.742T_1$), and $0.766T_1+0.5T_2=100$
Step 5. Plug-in the first equation into the second, we get $0.766T_1+0.5\times0.742T_1=100$ or $T_1\approx87.95lb$, and we also get $T_2\approx65.30lb$
Step 6. Rewrite the above vectors as: $\vec {T_1} =-56.5i+67.4j$ and $\vec {T_2} =56.5i+32.6j$.
