Question 524976
The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h.  If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel of the plane, and its ground speed.
<pre>
The other tutor used the law of cosines and the parallelogram method.
I will use the component method:

{{{drawing(400,400,-13,13,-13,13,
red(line(0,0,0,2),line(.3,1.7,0,2),line(-.3,1.7,0,2)),
triangle(-9,0,9,0,9,0),triangle(0,-9,0,0,0,0),triangle(0,2,0,9,0,9), locate(-10,.2,W),locate(9.5,.4,E), locate(-.3,-9,S),
locate(-.2,10,N), green(line(0,0,5.64510555,10.75120831),
line(5.64510555-1,10.75120831-1,5.64510555,10.75120831),
line(5.64510555+.05,10.75120831-1.3,5.64510555,10.75120831),
arc(0,0,16,-16,90-27.7,90)),locate(.4,7,"27.7°") 


) )}}}

The red vector represents the wind's velocity.  Since the wind
is FROM the SOUTH, it is blowing TOWARD the NORTH!  

The green vector represents the plane's velocity.
Note that that the angle 27.7° is measured clockwise from
the vertical (North), so the angle we will use is measured 
counterclockwise from the horizontal (East) calculated as 
90°-27.7° or 62.3°.

         x-components          y-components

Plane     255cos(90°)            255sin(90°)
Wind       42cos(62.3°)           42sin(62.3°)
----------------------------------------------
Sums:     19.52336592            292.1865323

|Resultant| = {{{sqrt(19.52336592^2 + 292.1865323^2)}}} = 292.8380635 mi/h

Angle (counterclockwise from East) = tan<sup>-1</sup>({{{292.1865323/19.52336592}}} = 86.177°

To find the bearing clockwise from the North, we subtract from 90° and get 3.823°,

and the bearing is written as N 3.823° E.

I agree with the other tutor on the magnitude of the resultant, 
but we differ just a bit on the angle.

Edwin</pre>