Answer
The magnitude of the resultant force is $270$ pounds and the angle of the resultant force is $27.7{}^\circ $.
Work Step by Step
A non-zero vector vector $\mathbf{v}$ is expressed as $\mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j}$ if the magnitude and direction angle from positive x-axis are $\left\| \mathbf{v} \right\|\text{ and }\theta $,
The direction angle for the first force ${{\mathbf{F}}_{\mathbf{1}}}$ in the positive direction of x-axis is $90{}^\circ -25{}^\circ =65{}^\circ $ and the direction angle for the second force ${{\mathbf{F}}_{\mathbf{2}}}$ in the positive direction of x-axis is $90{}^\circ -80{}^\circ =10{}^\circ $.
So, for the first force ${{\mathbf{F}}_{\mathbf{1}}}$, $\left\| {{\mathbf{F}}_{\mathbf{1}}} \right\|=100\text{ and }\theta =65{}^\circ $.
So,
$\begin{align}
& {{\mathbf{F}}_{\mathbf{1}}}=100\cos 65{}^\circ \mathbf{i}+100\sin 65{}^\circ \mathbf{j} \\
& =42.3\mathbf{i}+90.6\mathbf{j}
\end{align}$
Similarly, for the second force ${{\mathbf{F}}_{\mathbf{2}}}$, $\left\| {{\mathbf{F}}_{\mathbf{2}}} \right\|=200\text{ and }\theta =10{}^\circ $.
So,
$\begin{align}
& {{\mathbf{F}}_{\mathbf{2}}}=200\cos 10{}^\circ \mathbf{i}+200\sin 10{}^\circ \mathbf{j} \\
& =197\mathbf{i}+34.7\mathbf{j}
\end{align}$
Therefore, the resultant force $\mathbf{F}$ is ${{\mathbf{F}}_{\mathbf{1}}}+{{\mathbf{F}}_{\mathbf{2}}}$, that is,
$\begin{align}
& \mathbf{F}=\left( 42.3\mathbf{i}+90.6\mathbf{j} \right)+\left( 197\mathbf{i}+34.7\mathbf{j} \right) \\
& =\left( 42.3+197 \right)\mathbf{i}+\left( 90.6+34.7 \right)\mathbf{j} \\
& =239.3\mathbf{i}+125.3\mathbf{j}
\end{align}$
Magnitude of vector is given as below:
The magnitude of $\mathbf{r}=a\mathbf{i}+b\mathbf{j}$ is given by $\left\| \mathbf{r} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$.
Here, $\mathbf{F}=239.3\mathbf{i}+125.3\mathbf{j}$.
So,
$\begin{align}
& \left\| \mathbf{F} \right\|=\sqrt{{{239.3}^{2}}+{{125.3}^{2}}} \\
& \approx 270
\end{align}$
Now, to find the angle of the resultant force, we will use the following formula:
$\cos \theta =\frac{a}{\left\| \mathbf{F} \right\|}$ or $\sin \theta =\frac{b}{\left\| \mathbf{F} \right\|}$.
So,
$\begin{align}
& \cos \theta =\frac{a}{\left\| \mathbf{F} \right\|} \\
& =\frac{239.3}{270}
\end{align}$
This gives $\theta =27.7{}^\circ $.
So, the magnitude of the resultant force is $270$ pounds and the angle of the resultant force is $27.7{}^\circ $.
