Precalculus (6th Edition) Blitzer

The resultant force is $108.2\text{ pounds}$ and angle is $12.6{}^\circ$.
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, $\theta =12.6{}^\circ$