Answer
The speed of the plane is $\text{78 mph}$ and angle is $75.4{}^\circ $.
Work Step by Step
The force $\mathbf{F}$ in vector form is given as
$\mathbf{F}=\left\| \left. \mathbf{F} \right\| \right.\cos \theta \mathbf{i}+\left\| \left. \mathbf{F} \right\| \right.\sin \theta \mathbf{j}$
For $\text{N70}{}^\circ \text{E}$, the value of $\theta $ for the plane from x-axis is given as
$\begin{align}
& \theta =90{}^\circ -70{}^\circ \\
& =20{}^\circ
\end{align}$
Put the value $\theta =20{}^\circ $ and $\left\| \left. \mathbf{v} \right\| \right.=320$.
The vector $\mathbf{v}$ is given as
$\begin{align}
& \mathbf{v}=\left\| \left. \mathbf{v} \right\| \right.\cos 20{}^\circ \mathbf{i}+\left\| \left. \mathbf{v} \right\| \right.\sin 20{}^\circ \mathbf{j} \\
& =320\cos 20{}^\circ \mathbf{i}+320\sin 20{}^\circ \mathbf{j} \\
& =300.7\mathbf{i}+109.5\mathbf{j}
\end{align}$
The value of $\theta $ for ground speed of the plane is $30{}^\circ $.
Put the value $\theta =30{}^\circ $ and $\left\| \left. \mathbf{w} \right\| \right.=370$.
$\begin{align}
& \mathbf{w}=\left\| \left. \mathbf{w} \right\| \right.\cos 30{}^\circ \mathbf{i}+\left\| \left. \mathbf{w} \right\| \right.\sin 30{}^\circ \mathbf{j} \\
& =370\cos 30{}^\circ \mathbf{i}+370\sin 30{}^\circ \mathbf{j} \\
& =320.4\mathbf{i}+185\mathbf{j}
\end{align}$
The resultant vector is given by $\mathbf{w}-\mathbf{v}$.
Use the vector addition as follows:
$\begin{align}
& \mathbf{w}-\mathbf{v}=320.4\mathbf{i}+185\mathbf{j}-\left( 300.7\mathbf{i}+109.5\mathbf{j} \right) \\
& =19.7\mathbf{i}+75.5\mathbf{j}
\end{align}$
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 $19.7\mathbf{i}+75.5\mathbf{j}$.
Comparing it with the above equation gives
$\begin{align}
& a=19.7\mathbf{i} \\
& b=75.5\mathbf{j}
\end{align}$
Then magnitude of $\mathbf{w}-\mathbf{v}$ is given by $\left\| \mathbf{w}-\mathbf{v} \right\|$.
$\begin{align}
& \left\| \mathbf{w}-\mathbf{v} \right\|=\sqrt{{{\left( 19.7 \right)}^{2}}+{{\left( 75.5 \right)}^{2}}} \\
& \approx 78
\end{align}$
Then magnitude of $\mathbf{w}-\mathbf{v}$ is $\text{78}$.
The speed of the plane is $\text{78 mph}$.
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 $19.7\mathbf{i}+75.5\mathbf{j}$.
Then direction angle of $\mathbf{w}-\mathbf{v}$ is given by
$\begin{align}
& \cos \theta =\frac{19.7}{78} \\
& \theta ={{\cos }^{-1}}\left( \frac{19.7}{78} \right) \\
& \theta =75.4{}^\circ
\end{align}$
