Answer
The resultant force is $6353\text{ pounds}$ and angle is $1.7{}^\circ $.
Work Step by Step
Magnitude of the first force = $4200$ pounds.
The direction of the first force $=$ $\text{N}65{}^\circ \text{E}$.
Magnitude of the second force $=3000$ pounds.
The direction of the second force $=\text{S}58{}^\circ \text{E}$.
The vector component along the x-axis or horizontal is i and the vector component along the y-axis or the vertical direction is j. Let ${{F}_{1}}$ and ${{F}_{2}}$ denote the first and second force acting on the object, respectively. Therefore, the equation becomes
$\begin{align}
& {{F}_{1}}=4200\cos {{25}^{0}}\text{i}+4200\sin {{25}^{0}}\text{j} \\
& =3806.49\text{i}+1775\text{j} \\
& {{F}_{2}}=3000\cos {{328}^{0}}i+3000\sin {{328}^{0}}j \\
& =2544.14\text{i}-1589.76\text{j}
\end{align}$
And,
$\begin{align}
& F=\left( 3806.49+2544.14 \right)\text{i}+\left( 1775-1589.76 \right)\text{j} \\
& =6350.63\text{i}+185.24\text{j}
\end{align}$
So, the magnitude of force vector is as shown below
$\begin{align}
& \left| \left| F \right| \right|=\sqrt{{{\left( 6350.63 \right)}^{2}}+{{\left( 185.24 \right)}^{2}}} \\
& \approx 6353.33\text{ pounds}
\end{align}$
Also, the angle is
$\cos \theta =\frac{6350.63}{6353.33}$
Therefore,
$\theta =1.7{}^\circ $
