Answer
The resultant force is $108.2\text{ pounds}$ and angle is $12.6{}^\circ $.
Work Step by Step
Magnitude of the first force $=70\text{ pounds}$.
The direction of the first force $=$ $\text{S}56{}^\circ \text{E}$.
Magnitude of the second force $=50$ pounds.
The direction of the second force $=\text{N72}{}^\circ \text{E}$.
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
$\begin{align}
& {{F}_{1}}=70\cos 326{}^\circ \text{i}+70\sin {{326}^{0}}\text{j} \\
& =58\text{i}-\text{39}\text{.1j} \\
& {{F}_{2}}=50\cos {{18}^{0}}\text{i}+50\sin {{18}^{0}}\text{j} \\
& =47.6\text{i+}15.5\text{j}
\end{align}$
And,
$\begin{align}
& F=\left( 58+47.6 \right)\text{i}+\left( 15.5-39.1 \right)\text{j} \\
& =105.6\text{i}-23.6\text{j}
\end{align}$
So, the magnitude of force vector is as shown below
$\begin{align}
& \left| \left| F \right| \right|=\sqrt{{{\left( 105.6 \right)}^{2}}+{{\left( 23.6 \right)}^{2}}} \\
& =108.2\text{ pounds}
\end{align}$
Also, the angle is
$\begin{align}
& \text{cos}\theta =\frac{a}{\left| \left| F \right| \right|} \\
& \theta ={{\cos }^{-1}}\frac{105.6}{108.2} \\
& =12.6{}^\circ \\
& 360{}^\circ -12.6{}^\circ =347.4{}^\circ
\end{align}$
Therefore,
$\theta =12.6{}^\circ $
