Answer
$\begin{aligned} & F_R=5.93 \mathrm{kN} \\ & \theta=77.8^{\circ} P \\ & M_{R_A}=34.8 \mathrm{kN} \cdot \mathrm{m}\end{aligned}$
Work Step by Step
$
\begin{aligned}
\pm F_{R_s}=\Sigma F_x ; \quad F_{R_x} & =1.5 \sin 30^{\circ}-2.5\left(\frac{4}{5}\right) \\
& =-1.25 \mathrm{kN}=1.25 \mathrm{kN} \leftarrow \\
+\uparrow F_{R_y}=\Sigma F_y ; \quad F_{R_y} & =-1.5 \cos 30^{\circ}-2.5\left(\frac{3}{5}\right)-3 \\
& =-5.799 \mathrm{kN}=5.799 \mathrm{kN} \downarrow
\end{aligned}
$
Thus,
$
F_R=\sqrt{F_{R_x}^2+F_{R_s}^2}=\sqrt{1.25^2+5.799^2}=5.93 \mathrm{kN}
$$
and
$
\begin{gathered}
\theta=\tan ^{-1}\left(\frac{F_{R_y}}{F_{R_s}}\right)=\tan ^{-1}\left(\frac{5.799}{1.25}\right)=77.8^{\circ} P \\
\mathrm{↺}+M_{R_A}=\Sigma M_A ; \quad M_{R_A}=-2.5\left(\frac{3}{5}\right)(2)-1.5 \cos 30^{\circ}(6)-3(8) \\
=-34.8 \mathrm{kN} \cdot \mathrm{m}=34.8 \mathrm{kN} \cdot \mathrm{m} \quad(\text { Clockwise })
\end{gathered}
