SOLUTION: A girl starts from a point A and walks 285m to B on a bearing of 078degree. she then walks due south to a point C which is 307m from A. what is the bearing of A from C, and what is

Question 1065158: A girl starts from a point A and walks 285m to B on a bearing of 078degree. she then walks due south to a point C which is 307m from A. what is the bearing of A from C, and what is (BC)
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A girl starts from a point A and walks 285m to B on a bearing of 078degree. she then walks due south to a point C which is 307m from A. what is the bearing of A from C, and what is (BC)
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That's 2 sides of a triangle and the angle between them.
AB = 285 m (side c)
BC = 307 m (side a)
Angle B = 78 degs
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The 3rd side, side b, is the side AC
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Use the Cosine Law.
b^2 = a^2 + c^2 - 2bs*cos(B)
b^2 = 285^2 + 307^2 - 2*285*307*cos(78)
b^2 =~ 142538.29
b =~ 377.54 meters
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Use the Law of Sines to find angle C
c/sin(C) = b/sin(B)
285/sin(C) = b/sin(78)
sin(C) = 285*sin(78)/377.54
sin(C) =~ 0.738386
C =~47.6 degs
Bearing = 360 - C = 312� (only whole degrees are used in bearings)

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