If A is at (0,0) then
B is at (4 sin15°, 4 cos15°)
C is at B + (6 cos15°, -6 sin15°) =
(x,y) = (4 sin15° + 6 cos15° , 4 cos15° - 6 sin15°)
So the bearing of C from A is θ, where 90-θ where
tanθ = y/x
A man starts from a Point a and walks 4km on a bearing of 015 to point b. He then walks 6km on a bearing of 105 to c. What is the bearing of c from a
1 answer