Question
A man starts from a Point a and walks 4km on a bearing of 015. he then walks 6km on a bearing of 105 to c. What is the bearing of c from a.
Answers
Steve
he walks on a heading, not a bearing.
A-to-B = (4cos75°,4sin75°) = (1.03,3.86)
B-toC = (6cos15°,-6sin15°) = (5.80,-1.55)
so, A-to-C = (6.83,2.31)
Now,
tanθ = 2.31/6.83
θ = 18.7°
so, the bearing of C from A is 90-θ = 71.3°
A-to-B = (4cos75°,4sin75°) = (1.03,3.86)
B-toC = (6cos15°,-6sin15°) = (5.80,-1.55)
so, A-to-C = (6.83,2.31)
Now,
tanθ = 2.31/6.83
θ = 18.7°
so, the bearing of C from A is 90-θ = 71.3°