## Precalculus (6th Edition) Blitzer

The magnitude of the resultant force is $270$ pounds and the angle of the resultant force is $27.7{}^\circ$.
A non-zero vector vector $\mathbf{v}$ is expressed as $\mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j}$ if the magnitude and direction angle from positive x-axis are $\left\| \mathbf{v} \right\|\text{ and }\theta$, The direction angle for the first force ${{\mathbf{F}}_{\mathbf{1}}}$ in the positive direction of x-axis is $90{}^\circ -25{}^\circ =65{}^\circ$ and the direction angle for the second force ${{\mathbf{F}}_{\mathbf{2}}}$ in the positive direction of x-axis is $90{}^\circ -80{}^\circ =10{}^\circ$. So, for the first force ${{\mathbf{F}}_{\mathbf{1}}}$, $\left\| {{\mathbf{F}}_{\mathbf{1}}} \right\|=100\text{ and }\theta =65{}^\circ$. So, \begin{align} & {{\mathbf{F}}_{\mathbf{1}}}=100\cos 65{}^\circ \mathbf{i}+100\sin 65{}^\circ \mathbf{j} \\ & =42.3\mathbf{i}+90.6\mathbf{j} \end{align} Similarly, for the second force ${{\mathbf{F}}_{\mathbf{2}}}$, $\left\| {{\mathbf{F}}_{\mathbf{2}}} \right\|=200\text{ and }\theta =10{}^\circ$. So, \begin{align} & {{\mathbf{F}}_{\mathbf{2}}}=200\cos 10{}^\circ \mathbf{i}+200\sin 10{}^\circ \mathbf{j} \\ & =197\mathbf{i}+34.7\mathbf{j} \end{align} Therefore, the resultant force $\mathbf{F}$ is ${{\mathbf{F}}_{\mathbf{1}}}+{{\mathbf{F}}_{\mathbf{2}}}$, that is, \begin{align} & \mathbf{F}=\left( 42.3\mathbf{i}+90.6\mathbf{j} \right)+\left( 197\mathbf{i}+34.7\mathbf{j} \right) \\ & =\left( 42.3+197 \right)\mathbf{i}+\left( 90.6+34.7 \right)\mathbf{j} \\ & =239.3\mathbf{i}+125.3\mathbf{j} \end{align} Magnitude of vector is given as below: The magnitude of $\mathbf{r}=a\mathbf{i}+b\mathbf{j}$ is given by $\left\| \mathbf{r} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. Here, $\mathbf{F}=239.3\mathbf{i}+125.3\mathbf{j}$. So, \begin{align} & \left\| \mathbf{F} \right\|=\sqrt{{{239.3}^{2}}+{{125.3}^{2}}} \\ & \approx 270 \end{align} Now, to find the angle of the resultant force, we will use the following formula: $\cos \theta =\frac{a}{\left\| \mathbf{F} \right\|}$ or $\sin \theta =\frac{b}{\left\| \mathbf{F} \right\|}$. So, \begin{align} & \cos \theta =\frac{a}{\left\| \mathbf{F} \right\|} \\ & =\frac{239.3}{270} \end{align} This gives $\theta =27.7{}^\circ$. So, the magnitude of the resultant force is $270$ pounds and the angle of the resultant force is $27.7{}^\circ$.