True Velocity and Heading An airplane flies at $670$ MPH directly northwest. Wind blows at $70$ MPH from the west (i.e the wind is blowing towards the east). Determine the true velocity and heading of the plane.
Steps:
$<670\cos(3\pi/4), 670 \sin(3\pi/4)>$
$<(-670\sqrt2)/2), (670 \sqrt2)/2)>$
$<70,0>$
$622 \text{ MPH}$
$\arctan(-473.762/403.762)=-49.5^\circ $ rr $180-49.5=131.5^\circ$ North of East.
For this question off of a Multivariable textbook I am reading, here is the question and the answer. The problem is I don't understand anything that occurred for the steps. So if anyone could help me understand where $3\pi/4$, $622$ MPH, and $\arctan$ came from and help me to understand exactly how this problem was computed, that would be great!
 A: The Problem is basically a vector addition. You are given a magnitude V= 670MPH and a bearing (northwest, i.e $\theta$= 135 degrees measured from the horizontal i.e $\frac{3\pi}{4}$ radians measured from the horizontal). Now to convert a magnitude and a bearing to a vector you convert it into it's components:
$$(v_x , v_y)=(V\cos{\theta , V\sin{\theta})}$$
Where $v_x$ is the velocity in the x-direction (west) and $v_y$ is the velocity in the y-direction (north) . 
The reason you want to do this is because it is much easier to add vectors together and convert back to a magnitude and bearing than it is to just combine two scalars.
So back to the problem, using trigonometric relations:
$$(v_x , v_y)=(-670\frac{\sqrt2}{2} ,670 \frac{\sqrt2}{2} )$$
Now you are given the wind vector, note since it is blowing directly east it is already in the form of a vector: $(w_x , w_y)=(70,0)$ .
You can add the components of this vector together to give your the resultant velocity vector $(r_x ,r_y)=(v_x , v_y) +(w_x , w_y)=(-670\frac{\sqrt2}{2} +70,670 \frac{\sqrt2}{2} +0)\approx(-403.76 ,473.762)$ .
Now all thats left is to convert this back to a magnitude and bearing. To get the magniude (let's call it R):
$$R=\sqrt{r_x^2+r_y^2}\approx622 MPH$$
and bearing $\phi$ is calculated as:
$$\phi=\arctan{\frac{r_y}{r_x}}\approx-49.5 degrees$$ 
