Question 972501: A ship leaves port A and sails 150 km with heading 115 degrees to point B. It then changes its heading to 75 degrees and sails 120 km to C. Find its distance and bearing from A.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Need to draw this out, but I get a triangle with legs 115 and 120 (units all in km) with the missing length the distance between A and C.
The angle below the AC distance is 25 degrees from A. It has then changed 15 degrees to the north.. That makes angle ACB 140 degrees.
You have a 140 degree angle bounded by sides of 150 km and 115 km. The cosine will be negative there, so all terms will be positive.
b^2=a^2+c^2-2ac cos angle B
= (150)^2 + (120)^2- 2(150)(120) cos (140)
=22500+14400- 36000*(-.766)= 36900 + 27577.6=253.92 km, when one takes the square root.
The distance south of a due east line from A is
150 sin 25 or 63.39 km for the first leg
115*sin15 or 29.76 km for the second, which is the distance NORTH of point B.
So, the distance is 253.92 km with a southerly direction of (63.39-29.76)=33.63 km
The distance is the hypotenuse, and the distance south is the opposite leg from the angle southward from 90 degrees, or due east. Their quotient is 0.132, and the arc sin =7.6 degrees.
The bearing is 97.6 degrees, or just south of due east.
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