SOLUTION: Hi
could you kindly solve this trig question with a diagram please.
Town B is 40km due north of town A. The distance from B to C is 18km and the bearing of C from A
is 025 deg
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-> SOLUTION: Hi
could you kindly solve this trig question with a diagram please.
Town B is 40km due north of town A. The distance from B to C is 18km and the bearing of C from A
is 025 deg
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Question 1143501: Hi
could you kindly solve this trig question with a diagram please.
Town B is 40km due north of town A. The distance from B to C is 18km and the bearing of C from A
is 025 degrees. It is known that angle BCA is obtuse.
What is the bearing of C from B.
Thanks Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Town B is 40km due north of town A. The distance from B to C is 18km and the bearing of C from A
is 025 degrees. It is known that angle BCA is obtuse.
What is the bearing of C from B.
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Find angle C using the Law of Sines.
18/sin(25) = 40/sin(C)
sin(C) = 40sin(25)/18 =~ 0.93915
C =~ 110 degs (since it's spec's as obtuse)
Angle B = 180 - (A + C) = 45 degs
Bearing from C to B = 315 degs
