Answer
$\begin{aligned} & \left(M_{A B}\right)_A=3.88 \mathrm{kip} \cdot \mathrm{ft} \\ & \left(M_{B C D}\right)_A=2.05 \mathrm{kip} \cdot \mathrm{ft} \\ & \left(M_{\operatorname{man}}\right)_A=2.10 \mathrm{kip} \cdot \mathrm{ft}\end{aligned}$
Work Step by Step
Moment of the weight of boom $A B$ about point $A$ :
$
\begin{aligned}
↺+\left(M_{A B}\right)_A=-1500\left(10 \cos 75^{\circ}\right) & =-3882.29 \mathrm{lb} \cdot \mathrm{ft} \\
=3.88 \mathrm{kip} \cdot \mathrm{ft}(\text { Clockwise })
\end{aligned}
$
Moment of the weight of cage BCD about point $A$ :
$
\begin{aligned}
↺+\left(M_{B C D}\right)_A=-200\left(30 \cos 75^{\circ}+2.5\right) & =-2052.91 \mathrm{lb} \cdot \mathrm{ft} \\
=2.05 \mathrm{kip} \cdot \mathrm{ft}(\text { Clockwise })
\end{aligned}
$
Moment of the weight of the man about point A:
$
\begin{aligned}
\mathrm{↺}+\left(M_{\max }\right)_A=-175\left(30 \cos 75^{\circ}+4.25\right) & =-2102.55 \mathrm{lb} \cdot \mathrm{ft} \\
=2.10 \mathrm{kip} \cdot \mathrm{ft}(\text { Clockwise })
\end{aligned}
$
