## Precalculus (6th Edition) Blitzer

The speed of the plane is $83\text{ mph}$ and angle is $68.8{}^\circ$.
The force $\mathbf{F}$ in vector form is given as $\mathbf{F}=\left\| \left. \mathbf{F} \right\| \right.\cos \theta \mathbf{i}+\left\| \left. \mathbf{F} \right\| \right.\sin \theta \mathbf{j}$ Fordirection $\text{S36}{}^\circ \text{E}$, the value of $\theta$ for the plane from x-axis is given as \begin{align} & \theta =270{}^\circ +36{}^\circ \\ & =306{}^\circ \end{align} Put the value $\theta =306{}^\circ$ and $\left\| \left. \mathbf{v} \right\| \right.=540$. The vector $\mathbf{v}$ is given as \begin{align} & \mathbf{v}=\left\| \left. \mathbf{v} \right\| \right.\cos 306{}^\circ \mathbf{i}+\left\| \left. \mathbf{v} \right\| \right.\sin 306{}^\circ \mathbf{j} \\ & =540\cos 306{}^\circ \mathbf{i}+540\sin 306{}^\circ \mathbf{j} \\ & =317.4\mathbf{i}-436.9\mathbf{j} \end{align} Fordirection $\text{S44}{}^\circ \text{E}$, the value of $\theta$ for the plane from x-axis is given as \begin{align} & \theta =270{}^\circ +44{}^\circ \\ & =314{}^\circ \end{align} The value of $\theta$ for ground speed of the plane is $314{}^\circ$. Put the value $\theta =314{}^\circ$ and $\left\| \left. \mathbf{w} \right\| \right.=500$. \begin{align} & \mathbf{w}=\left\| \left. \mathbf{w} \right\| \right.\cos 314{}^\circ \mathbf{i}+\left\| \left. \mathbf{w} \right\| \right.\sin 314{}^\circ \mathbf{j} \\ & =500\cos 314{}^\circ \mathbf{i}+500\sin 314{}^\circ \mathbf{j} \\ & =347.3\mathbf{i}-359.7\mathbf{j} \end{align} The resultant vector is given by $\mathbf{w}-\mathbf{v}$. Use the vector addition as follows \begin{align} & \mathbf{w}-\mathbf{v}=347.3\mathbf{i}-359.7\mathbf{j}-\left( 317.4\mathbf{i}-436.9\mathbf{j} \right) \\ & =29.9\mathbf{i}+77.2\mathbf{j} \end{align} If the vector is $\mathbf{F}=a\mathbf{i}+b\mathbf{j}$, then magnitude of $\mathbf{F}$ is given by $\left\| \mathbf{F} \right\|$. $\left\| \mathbf{F} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ The vector $\mathbf{w}-\mathbf{v}$ is $29.9\mathbf{i}+77.2\mathbf{j}$. Comparing it with the above equation gives \begin{align} & a=29.9\mathbf{i} \\ & b=77.2\mathbf{j} \end{align} Then magnitude of $\mathbf{w}-\mathbf{v}$ is given by $\left\| \mathbf{w}-\mathbf{v} \right\|$. \begin{align} & \left\| \mathbf{w}-\mathbf{v} \right\|=\sqrt{{{\left( 29.9 \right)}^{2}}+{{\left( 77.2 \right)}^{2}}} \\ & \approx 82.8 \\ & \approx 83 \end{align} Then magnitude of $\mathbf{w}-\mathbf{v}$ is $83$. The speed of the plane is $83\text{ mph}$. If the vector is $\mathbf{F}=a\mathbf{i}+b\mathbf{j}$, then magnitude of $\mathbf{F}$ is given by $\left\| \mathbf{F} \right\|$. $\left\| \mathbf{F} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ And the direction angle is given as $\cos \theta =\frac{a}{\left\| \mathbf{F} \right\|}$ The vector $\mathbf{w}-\mathbf{v}$ is $29.9\mathbf{i}+77.2\mathbf{j}$. Then direction angle of $\mathbf{w}-\mathbf{v}$ is given by \begin{align} & \cos \theta =\frac{a}{\left\| \mathbf{w}-\mathbf{v} \right\|} \\ & \cos \theta =\frac{29.9}{82.8} \\ & \theta ={{\cos }^{-1}}\left( \frac{29.9}{82.8} \right) \\ & \theta =68.8{}^\circ \end{align}