Answer
The magnitude is $\text{323 pounds}$ and the direction of the resultant force is $\text{E3}\text{.4 }\!\!{}^\circ\!\!\text{ N}$.
Work Step by Step
Here,
${{\text{F}}_{\text{1}}}=\text{250 pounds, N60 }\!\!{}^\circ\!\!\text{ E and }{{\text{F}}_{\text{2}}}=\text{150 pounds, S45 }\!\!{}^\circ\!\!\text{ E}$
The standard equation used to denote the position of force $ F\left( r,\theta \right)$ is:
$ F\left( r,\theta \right)=F\text{cos}\theta \mathbf{i}+\text{Fsin}\theta \mathbf{j}$ …… (1)
${{\text{F}}_{\text{1}}}$ makes a $90{}^\circ -60{}^\circ =30{}^\circ $ angle with $\text{F}$.
Then, ${{\text{F}}_{\text{1}}}$ is denoted in equation (1) as ${{\text{F}}_{\text{1}}}=\text{250 pounds}$
$\begin{align}
& {{\text{F}}_{1}}\left( 250,30{}^\circ \right)=250\text{cos}30\text{ }\!\!{}^\circ\!\!\text{ }\mathbf{i}+\text{250sin30}{}^\circ \mathbf{j} \\
& \text{=250}\left( \frac{\sqrt{3}}{2} \right)\mathbf{i}+250\left( \frac{1}{2} \right)\mathbf{j} \\
& \text{=216}\text{.5}\mathbf{i}+\text{125}\mathbf{j}
\end{align}$
Now, ${{\text{F}}_{2}}$ makes a $360{}^\circ -45{}^\circ =315{}^\circ $ angle with $\text{F}$.
Then ${{\text{F}}_{2}}$ is denoted in equation (1) as ${{\text{F}}_{2}}\text{=150 pounds}$
$\begin{align}
& {{\text{F}}_{2}}\left( 150,315{}^\circ \right)=150\text{cos315 }\!\!{}^\circ\!\!\text{ }\mathbf{i}+\text{150sin315}{}^\circ \mathbf{j} \\
& \text{=150}\left( \frac{1}{\sqrt{2}} \right)\mathbf{i}+250\left( -\frac{1}{\sqrt{2}} \right)\mathbf{j} \\
& \text{=106}\mathbf{i}-\text{106}\mathbf{j}
\end{align}$
Since the resultant force $ F\left( r,\theta \right)$ is given by,
$\begin{align}
& F\left( r,\theta \right)={{\text{F}}_{1}}\left( r,\theta \right)+{{\text{F}}_{2}}\left( r,\theta \right) \\
& \text{=}\left( \text{216}\text{.5}\mathbf{i}+\text{125}\mathbf{j} \right)+\left( \text{106}\mathbf{i}-\text{106}\mathbf{j} \right) \\
& \text{=322}\text{.5}\mathbf{i}+\text{19}\mathbf{j}
\end{align}$
Then,
$\begin{align}
& \text{F}\left( r,\theta \right)=\sqrt{{{322.5}^{2}}+{{19}^{2}}} \\
& \approx \text{323}\,\text{pounds}
\end{align}$
And, the angle is represented as below:
$\begin{align}
& \cos \theta =\frac{322.5}{323} \\
& \theta ={{\cos }^{-1}}\left( \frac{322.5}{323} \right) \\
& =3.4{}^\circ
\end{align}$
