Expert: Steve Holleran - 12/31/2007

Question
I am having trouble with these set of questions.

A plane is heading west at 200 mph. The wind is blowing S30°W at 25 mph.

What is the ground speed of the plane?

When you solve the triangle, what is the smallest angle?

What is the largest angle of the triangle?

What is the remaining angle?

What is the bearing of the plane?  

Answer
Hello Sana,

Okay, if you draw a North-South-East-West system, then the plane is heading due West at 200 mph.  If the wind vector is 30 degrees West of South, draw this from the origin, the "slide" this vector so it is tail-to-head with the plane's vector.

The points would look something like this:


          A                          O



     B

Where O is the origin, A is the plane's position, and B is the endpoint for the wind vector.  Connect them with lines and you'll see what I mean.

Now, the angle at A is 120 degrees  (90 + 30), the length OA is 200, the length AB is 25, so to find the plane's ground speed, you want BO = x.  You can set up the law of cosines:

 x^2 = 25^2 + 200^2 - 2 * 25 * 200 * cos 120

Now cos 120 = -1/2, so you get

 x^2 = 625 + 40000 + 5000 = 45,625

so x = 213.6  = ground speed of plane.

Then use the law of sines for the second angle:

 213.6/ sin 120  = 25 / sin O  = 200 / sin B

Solving for the angle at O, you get 5.8 degrees.  This is the smallest angle.  The largest is the 120, and the remaining angle is 54.2 degrees.

Since the angle at O is 5.8 degrees the bearing (degrees clockwise from North) is 270 - 5.8 = 264.2


Hope this is what you needed.

Steve