Precalculus (6th Edition) Blitzer

The resultant force is $85.18\text{ kg}$ and the angle is $162.9{}^\circ$. Magnitude of the first force is $64\text{ kg}$.
The direction of the first force is $\text{N39 }\!\!{}^\circ\!\!\text{ W}$. Magnitude of the second force is $48\text{ kg}$. The direction of the second force is $\text{S59 }\!\!{}^\circ\!\!\text{ W}$. 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}}=64\cos 129{}^\circ \mathbf{i}+64\sin 129{}^\circ \mathbf{j} \\ & =-40.28\mathbf{i}+49.74\mathbf{j} \\ & {{F}_{2}}=48\cos 211{}^\circ \mathbf{i}+48\sin 211{}^\circ \mathbf{j} \\ & =-41.14\mathbf{i}-24.72\mathbf{j} \end{align} And, \begin{align} & F=\left( -40.28-41.14 \right)\mathbf{i}+\left( 49.74-24.72 \right)\mathbf{j} \\ & =-81.42\mathbf{i}+25.02\mathbf{j} \end{align} So, the magnitude of force vector is as shown below \begin{align} & \left| \left| F \right| \right|=\sqrt{{{\left( -81.42 \right)}^{2}}+{{\left( 25.02 \right)}^{2}}} \\ & =85.18 \end{align} Also, the angle is \begin{align} & \text{cos}\theta =\frac{-81.42}{85.18} \\ & =162.9{}^\circ \end{align} Therefore, $\theta =162.9{}^\circ$