**Solution:**

**Approach-**

Given those, the magnitude and angle of action of R fallow from Equation,

$
\theta=\tan ^{-1}\left(\frac{R_y}{R_x}\right) \\
$

**Solution:-**

The components of the $8001 \mathrm{~b}$ force are,

$
\begin{aligned} \\
F_{x, 1} &=(800 \mathrm{lb}) \cos 45^{\circ} \\\\
&=56.7 \mathrm{lb} \\\\
F_{y, 1} &=(800 \mathrm{~J}) \sin 45^{\circ} \\\\
&=565.7 \mathrm{lb}\\
\end{aligned}\\
$

and $F_1$ is written in vector from as,

$
F_1=565.7 j+565 l 7 j 13\\
$

By Using the same procedure for the other two forces,

$
\begin{aligned}\\
F_2 &=-\left(350 \sin 20^{\circ}\right) j+\left(350 \cos 20^{\circ}\right) j \mathrm{db} \\\\
&=-119.7 j+328.9 j \mathrm{Jb} \\\\
F_3 &=-150 i \mathrm{Jb}\\
\end{aligned}\\
$

To calculate the components of the resultant, the horizontal and,

$
\begin{aligned} \\
R_x &=565.7-119.7-150 \mathrm{lb} \\\\
&=296.0 \mathrm{lb} \\\\
R_y &=565.7+328.9 \mathrm{lb} \\\\
&=894.9 \mathrm{lb}\\
\end{aligned}\\
$

The magnitude of the resultant force is

$
\begin{aligned}\\
R &=\sqrt{(296.0 \mathrm{Jb})^2+(894.6 \mathrm{Jb})^2} \\\\
&=942.3 \mathrm{lb}\\
\end{aligned}\\
$

and it acts at the angle

$
\begin{aligned}\\
a &=\tan ^{-1}\left(\frac{894.6 \lambda_b}{296.0 \lambda_b}\right) \\\\
&=\tan ^{-1}(3.022) \\\\
\Delta &=71.69^{\circ}\\
\end{aligned}\\
$

Discussion :-

The assistant force is larger than any one force, since $R_x$ and $R_y$ are positive,

The tip of the resultant vector lies in the first quadrant of the $x-y$ Plame.

$
\begin{aligned}\\
&R=942.3 \mathrm{lb} \\\\
&\partial=71.69^{\circ}\\
\end{aligned}\\
$