Answer
$$
\begin{aligned}
& M_O=4.27 \mathrm{~N} \cdot \mathrm{m} \\
& \alpha=95.2^{\circ} \\
& \beta=110^{\circ} \\
& \gamma=20.6^{\circ}
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
& \mathbf{r}_A=0.2 \sin 15^{\circ} \mathbf{i}+0.2 \cos 15^{\circ} \mathbf{j}+0.075 \mathbf{k} \\
& =0.05176 \mathbf{i}+0.1932 \mathbf{j}+0.075 \mathbf{k} \\
& \mathbf{F}=-20 \cos 15^{\circ} \mathbf{i}+20 \sin 15^{\circ} \mathbf{j} \\
& =-19.32 \mathbf{i}+5.176 \mathbf{j} \\
& M_O=\mathbf{r}_A \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0.05176 & 0.1932 & 0.075 \\
-19.32 & 5.176 & 0
\end{array}\right| \\
& =\{-0.3882 \mathbf{i}-1.449 \mathbf{j}+4.00 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m} \\
& M_O=4.272=4.27 \mathrm{~N} \cdot \mathrm{m} \\
& \alpha=\cos ^{-1}\left(\frac{-0.3882}{4.272}\right)=95.2^{\circ} \\
& \beta=\cos ^{-1}\left(\frac{-1.449}{4.272}\right)=110^{\circ} \\
& \gamma=\cos ^{-1}\left(\frac{4}{4.272}\right)=20.6^{\circ} \\
&
\end{aligned}
$$
