Precalculus (6th Edition) Blitzer

The magnitude and compass direction of the resultant force are $4232.08\ \text{pounds}$ and $72.68{}^\circ$, respectively.
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 $\left( 90{}^\circ -17.32{}^\circ \right)=72.68{}^\circ$.