Answer
$\begin{aligned} & T_{H A}=294 \mathrm{~N} \\ & T_{A B}=340 \mathrm{~N} \\ & T_{A E}=170 \mathrm{~N} \\ & T_{B D}=490 \mathrm{~N} \\ & T_{B C}=562 \mathrm{~N}\end{aligned}$
Work Step by Step
At $H$ :
$
\begin{array}{cl}
+\uparrow \Sigma F_y=0 ; & T_{H A}-30(9.81)=0 \\
& T_{H A}=294 \mathrm{~N}
\end{array}
$
At $A$ :
$
\begin{array}{ll}
+\uparrow \Sigma F_y=0 ; & T_{A B} \sin 60^{\circ}-30(9.81)=0 \\
& T_{A B}=339.83=340 \mathrm{~N} \\
\pm \Sigma F_x=0 ; & T_{A E}-339.83 \cos 60^{\circ}=0 \\
& T_{A E}=170 \mathrm{~N}
\end{array}
$
At $B$ :
$
\begin{array}{ll}
+\uparrow \Sigma F_y=0 ; & T_{B D}\left(\frac{3}{5}\right)-339.83 \sin 60^{\circ}=0 \\
& T_{B D}=490.50=490 \mathrm{~N} \\
\stackrel{\mathrm{t}}{\rightarrow} F_x=0 ; & 490.50\left(\frac{4}{5}\right)+339.83 \cos 60^{\circ}-T_{B C}=0 \\
& T_{B C}=562 \mathrm{~N}
\end{array}
$
