Trig
posted by Mary .
This has to do with vectors...I'm wondering if I set up the diagram correctly?
A plane has a speed of 250mph and flies with a bearing of N35degE. The wind is blowing from the west at 25 mph with a heading of due east. What is the actual speed and heading of the plane?

Trig 
Mary
Whoops I forgot to include the diagram...here it is:
i.imgur [dot] com/RRY5pox.png 
Trig 
Reiny
I would tack on the wind vector to the end of the plane vector at the point P so the final end point is Q
Join Q to the origin O
vector OQ has our required direction and its magnitude represents the actual speed
By simple geometry angle P = 125°
so by the cosine law:
OP^2 = 250^2 + 25^2  2(250)(25)cos 125°
= 70294.705..
OP = 265.13
So the actual speed is 265.13 mph
Let the additional angle be Ø
by the sine law:
sinØ/25 = sin 125/265.13
sinØ = .07724
Ø = 4.43°
Ø + 35° = 39.43°
So the heading should be N 39.43° E 
Trig 
Mary
Thank you <3 you are a lifesaver!
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