Answer
$\vec {T_1} \approx-14115i+5783j$ and $\vec {T_2} \approx14115i+12495j$.
Work Step by Step
Step 1. Identify the vector for the weight as $\vec {W} =-18278j$
Step 2. Identify the vector for T1 as $\vec {T_1} =-T_1cos22.3^{\circ} i +T_1sin22.3^{\circ} j\approx-0.925T_1i +0.379T_1 j$
Step 3. Identify the vector for T2 as $\vec {T_2} =T_2cos41.5^{\circ} i +T_2sin41.5^{\circ} j\approx0.749T_2i +0.663T_2 j$
Step 4. For a balanced force, the total should be zero which gives $\vec {W} +\vec {T_1} +\vec {T_2} =0$ and this leads to $0.749T_2=0.925T_1$ (or $T_2=1.235T_1$) and $0.379T_1+0.663T_2=18278$
Step 5. Plug-in the first equation into the second one, we get $0.379T_1+0.663\times1.235T_1=18278$ which gives $T_1=15259.58lb$, and we also get $T_2=18845.58lb$
Step 6. Rewrite the above vectors as: $\vec {T_1} \approx-14115i+5783j$ and $\vec {T_2} \approx14115i+12495j$.
