Answer
a) The vectors $\mathbf{v}$ and $\mathbf{w}$ are given as $\mathbf{v}=137.88\mathbf{i}+115.7\mathbf{j},\ \mathbf{w}=40\mathbf{i}$.
b) The resultant vector $\mathbf{v}+\mathbf{w}$ is $177.88\mathbf{i}+115.7\mathbf{j}$.
c) Approximate ground speed of plane is $212\text{ mph}$.
d) The plane true bearing is $\theta =\text{N}57{}^\circ \text{E}\text{.}$
Work Step by Step
(a)
The force $\mathbf{F}$ in vector form is given as
$\mathbf{F}=\left\| \left. \mathbf{F} \right\| \right.\cos \theta \mathbf{i}+\left\| \left. \mathbf{F} \right\| \right.\sin \theta \mathbf{j}$
The value of $\theta $ from x-axis is given as
$\begin{align}
& \theta =90{}^\circ -50{}^\circ \\
& =40{}^\circ
\end{align}$
Put the value $\theta =40{}^\circ $ and $\left\| \left. \mathbf{v} \right\| \right.=180$.
The vector $\mathbf{v}$ is given as
$\begin{align}
& \mathbf{v}=\left\| \left. \mathbf{v} \right\| \right.\cos 40{}^\circ \mathbf{i}+\left\| \left. \mathbf{v} \right\| \right.\sin 40{}^\circ \mathbf{j} \\
& =180\cos 40{}^\circ \mathbf{i}+180\sin 40{}^\circ \mathbf{j} \\
& =137.88\mathbf{i}+115.7\mathbf{j}
\end{align}$
Put the value $\theta =0{}^\circ $ and $\left\| \left. \mathbf{w} \right\| \right.=40$.
$\begin{align}
& \mathbf{w}=\left\| \left. \mathbf{w} \right\| \right.\cos 0{}^\circ \mathbf{i}+\left\| \left. \mathbf{w} \right\| \right.\sin 0{}^\circ \mathbf{j} \\
& =40\mathbf{i}
\end{align}$
(b)
Use the vector addition as follows:
$\begin{align}
& \mathbf{v}+\mathbf{w}=137.88\mathbf{i}+115.7\mathbf{j}+40\mathbf{i} \\
& =177.88\mathbf{i}+115.7\mathbf{j}
\end{align}$
(c)
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 $\mathbf{v}+\mathbf{w}$ is $177.88\mathbf{i}+115.7\mathbf{j}$.
Comparing it with the above equation gives
$\begin{align}
& a=177.88 \\
& b=115.7
\end{align}$
Then magnitude of $\mathbf{v}+\mathbf{w}$ is given by $\left\| \mathbf{v}+\mathbf{w} \right\|$.
$\begin{align}
& \left\| \mathbf{v}+\mathbf{w} \right\|=\sqrt{{{\left( 177.88 \right)}^{2}}+{{\left( 115.7 \right)}^{2}}} \\
& \approx 212
\end{align}$
Then magnitude of $\mathbf{v}+\mathbf{w}$ is $212$.
(d)
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 $\mathbf{v}+\mathbf{w}$ is $177.88\mathbf{i}+115.7\mathbf{j}$.
Then direction angle of $\mathbf{v}+\mathbf{w}$ is given by
$\begin{align}
& \cos \theta =\frac{177.88}{212} \\
& \theta ={{\cos }^{-1}}\left( \frac{177.88}{212} \right) \\
& \theta =33{}^\circ
\end{align}$
The direction angle is given by
$90{}^\circ -33{}^\circ =57{}^\circ $
