Answer
The speed of the plane is $83\text{ mph}$ and angle is $68.8{}^\circ $.
Work Step by Step
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
$\begin{align}
& \cos \theta =\frac{a}{\left\| \mathbf{w}-\mathbf{v} \right\|} \\
& \cos \theta =\frac{29.9}{82.8} \\
& \theta ={{\cos }^{-1}}\left( \frac{29.9}{82.8} \right) \\
& \theta =68.8{}^\circ
\end{align}$
