Answer
The magnitude and compass direction of the resultant force are $4232.08\ \text{pounds}$ and $72.68{}^\circ $, respectively.
Work Step by Step
Let the one tug pull with a force ${{\mathbf{F}}_{1}}$.
The second tug pulls with a force ${{\mathbf{F}}_{2}}$.
The one tug pulls with a force
$\begin{align}
& {{\mathbf{F}}_{1}}=\left\| {{\mathbf{F}}_{1}} \right\|\cos \theta \ \mathbf{i}+\left\| {{\mathbf{F}}_{1}} \right\|\sin \theta \ \mathbf{j} \\
& =2500\cos 35{}^\circ \mathbf{i}+2500\sin 35{}^\circ \mathbf{j} \\
& =2047.88\mathbf{i}+1433.94\mathbf{j}
\end{align}$
Here, $\theta $ is the angle of first force from the horizontal axis.
The second tug pulls with a force
$\begin{align}
& {{\mathbf{F}}_{2}}=\left\| {{\mathbf{F}}_{2}} \right\|\cos \varphi \ \mathbf{i}+\left\| {{\mathbf{F}}_{2}} \right\|\sin \varphi \ \mathbf{j} \\
& =2000\cos \left( -5{}^\circ \right)\mathbf{i}+2000\sin \left( -5{}^\circ \right)\mathbf{j} \\
& =1992.4\mathbf{i}-174.31\mathbf{j}
\end{align}$
Here, $\varphi $ is the angle of the second force from the horizontal axis.
The resultant force is
$\begin{align}
& \mathbf{F}={{\mathbf{F}}_{1}}+{{\mathbf{F}}_{2}} \\
& =\left( 2047.88\mathbf{i}+1433.94\mathbf{j} \right)+\left( 1992.4\mathbf{i}-174.31\mathbf{j} \right) \\
& =4040.28\mathbf{i}+1259.63\mathbf{j}
\end{align}$
The magnitude of the resultant force is
$\begin{align}
& \mathbf{F}=\sqrt{{{\left( 4040.28 \right)}^{2}}+{{\left( 1259.63 \right)}^{2}}} \\
& =4232.08\ \text{pounds}
\end{align}$
The direction of the resultant force is
$\begin{align}
& \tan \phi =\frac{1259.63}{4040.28} \\
& \phi =17.32{}^\circ
\end{align}$
Here, $\phi $ is the angle of the resultant force from the horizontal axis.
So, the compass direction of the resultant force is $\left( 90{}^\circ -17.32{}^\circ \right)=72.68{}^\circ $.
