Answer
The resultant force is $2038\text{ kg}$ and the angle is $162.8{}^\circ $.
Work Step by Step
Magnitude of the first force is $1610\text{ kg}$.
The direction of the first forceis $\text{N35 }\!\!{}^\circ\!\!\text{ E}$.
Magnitude of the second forceis $\text{1250 kg}$.
The direction of the second forceis $\text{S55 }\!\!{}^\circ\!\!\text{ W}$.
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
$\begin{align}
& {{F}_{1}}=1610\cos {{125}^{0}}\mathbf{i}+1610\sin {{125}^{0}}\mathbf{j} \\
& =-923.46\mathbf{i}+1318.83\mathbf{j} \\
& {{F}_{2}}=1250\cos {{215}^{0}}\mathbf{i}+1250\sin {{215}^{0}}\mathbf{j} \\
& =-1023.94\mathbf{i}-716.97\mathbf{j}
\end{align}$
And,
$\begin{align}
& F=\left( -923.46-1023.94 \right)\mathbf{i}+\left( 1318.83-716.97 \right)\mathbf{j} \\
& =-1947.40\mathbf{i}+601.86\mathbf{j}
\end{align}$
So, the magnitude of force vector is as shown below
$\begin{align}
& \left| \left| F \right| \right|=\sqrt{{{\left( -1947.40 \right)}^{2}}+{{\left( 601.86 \right)}^{2}}} \\
& \approx 2038.28
\end{align}$
Also, the angle is
$\cos \theta =\frac{-1947.40}{2038.28}$
Therefore,
$\theta =162.8{}^\circ $
