Posted by miles on .
What heading (or direction) and airspeed (speed in still air) are required for an airplane to fly 837 mph due north if a wind of 63 mph is blowing in the direction of S 11.5 degrees E?

calculus iii 
Reiny,
The information translated into triangle ABC,
where BC is vertical and magnitude 837
AB is our required vector
and angle C = 168.5°
by cosine law
AB^2 = 837^2 + 63^2  2(837)(63)cos 168.5°
AB = 898.82
by sine law:
Sin B / 63 = sin168.5/898.82
sinB = .013974
B = .8 °