Answer
$\begin{aligned} & F_{A B}=50 \mathrm{lb} \\ & F_{A D}=70.7 \mathrm{lb} \\ & h=23 \mathrm{ft}\end{aligned}$
Work Step by Step
At point $B$ :
$
\begin{array}{ll}
+\uparrow \Sigma F_y=0 ; & \frac{1}{\sqrt{2}} F_{B C}-50=0 \\
& F_{B C}=70.71=70.7 \mathrm{lb} \\
\Sigma F_X=0 ; & \frac{1}{\sqrt{2}}(70.71)-F_{A B}=0 \\
& F_{A B}=50 \mathrm{lb}
\end{array}
$
At point $A$ :
$
\begin{array}{ll}
\stackrel{ \pm}{\rightarrow} F_x=0 ; & 50-F_{A D} \cos \theta=0 \\
+\uparrow \Sigma F_y=0 ; & F_{A D} \sin \theta-50=0 \\
& \theta=45^{\circ} \\
& F_{A D}=70.7 \mathrm{lb} \\
& h=18+5=23 \mathrm{ft}
\end{array}
$
