An airplane is flying from Montreal to Vancouver. The wind is blowing from the west at 60km/hour. The airplane flies at an airspeed of 750km/h and must stay on a heading of 65 degrees west of north.

Question

A) what heading should the pilot take to compensate for the wind?
B) what is the speed of the airplane relative to the ground?

We are suppose to use the cosine law and I tried to draw a diagram which formed a triangle. But I have no idea how to go on or what to do

Answer 1

I don't know what your diagram looks like, but try this
I will assume that this is a question dealing with vectors.
let O be the centre of you NS, EW grid
draw a line OB with a heading of N65°W , and mark it x km/h
this will be your resultant vector
draw in your wind vector AB as a horizontal line and mark it 60 km/h
Join AO and mark it 750 km/h
In triangle OBA, angle ABO = 155°
let angle AOB = Ø
by the sine law,
sinØ/60 = sin155/750
Ø = 1.94°

so he should head in the direction of N 66.94° W

then angle A = 23.06°

x/sin23.06 = 750/sin155
x = 695.19 km/h

using the cosine law would be more difficult,
e.g.
750^2 = x^2 + 60^2 - 2(x)(60)cos155
558900 = x^2 + 108.7569..x
x^2 + 108.7569x - 558900 = 0
using the quadratic formula ...
x = 695.19 as above or x = a negative, which must be rejected.
Then you would use the sine law to find the angle

Answer 2

If the apparent heading of the plane is (65+θ)°, and the plane's speed is v, then

60+v*sin65° = 750 sin(65+θ)°
750 cos(65+θ)° = v cos65°

v = 695 and θ=1.6°
So, the plane needs to fly at N66.6°W
and the plane's ground speed is 695 km/hr

Not sure how to use the law of cosines for this one.