Question 1105606: A plane flies 710 miles from A to B with a bearing of N 65 degrees east. Then it flies 500 miles from B to C with a bearing of N 39 degrees east. Find the straight line distance and bearing from C to A.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! If one draws this out, it is an obtuse triangle with angle B being 154 degrees and the legs on either side being 500 (a) and 710 (c)
Draw a line from B down 90 degrees (east) of A and call that point N. The angle from B is 65 degrees, since angle BAN is 25 degrees and ANB is 90 degrees.
From B, the angles are 65 degrees, 90 degrees, 51 degrees (complement of the 39) and angle B itself, which is 360-(206)=154 degrees.
Law of Cosines: b^2=a^2+c^2-2ac cos B
b^2=500^2+710^2-2(500)(710) cos 154
=754100-710000*(-0.8988)
b=1179.93 miles (not b^2)
The bearing needs Law of Sines
sin A/500=sin B/1179.93=0.4384/1179.93
sin A=500(0.4384)/1179.93
angle A=10.71 degrees.
But the 90 degree quadrant includes the bearing from N to east PLUS 10.71 degrees+25 degrees angle BAN.
90-35.71 degrees or 54.29 degrees
The bearing is from C to A, so this is the reciprocal or 234.29 degrees.
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