SOLUTION: An aircraft fliesrounda triangular course. The first leg is 200km on a bearing of 115^0 and the second leg is 150km on a bearing of 230^0. How long is the third leg of the course a

Question 1070579: An aircraft fliesrounda triangular course. The first leg is 200km on a bearing of 115^0 and the second leg is 150km on a bearing of 230^0. How long is the third leg of the course and what bearing must the aircraft fly?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Draw this.
The SE corner angle is 115 degrees, and we will call that C.
the third leg will be c
c^2=a^2+b^2-2ab cos C
=40000+22500-2(200)(150)cos 115
=40000+22500+25357, the cosine of 115 being negative
c^2=87857.10
c=296.41 km, consistent with continuing to move away from the starting point on the second part.
For the angles, use the law of sines
sin 115/296.41=sin x/150
sin 115*150/296.41=0.4586
arc sin gives 27.3 degrees, this being the angle between the first and third legs.
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last one is sin y/200=sin 115/296.41
sin y=200*sin 115/296.41
sin y=0.6115
arc sin gives 37.7 degrees
The angles add to 180 degrees.
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The angle from the second leg to the final leg is 37.7 degrees
The reciprocal bearing is 50 degrees, and the angle is taken to the left of that or northward, so that the bearing is 12.3 degrees.
12.3 degree bearing for 296.41 km

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