A girl starts from a point A and walks 285m to B on a bearing of 078°, 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 1

If A is at (0,0), then
B is at (285sin78°,285cos78°) = (278.77,59.25)
Now we know that C is at (278.77,59.25-c) where
278.77^2 + (59.25-c)^2 = 307^2
c = 187.84
So, C is at (278.77,-128.59)
So, in triangle ABC, using the law of cosines,
285^2 = 187.84^2 + 307^2 - 2*187.84*307*cosA
A = 65°
Thus the bearing of A from C is 360-A = 295°
As noted above, BC = 187.84

Answer 2

All angles are measured CW from +y-axis.
.d = 285*sin78 = 278.8 m. = hor. distance of AB from Y-axis.
h = 285*Cos78 = 59.3 m = ht. of AB.

sinX = 278.8/307.
X = 65o E. of S. = 115o CW. = Bearing of AC.

BC = 59.3 + 307*sin(90-65) = 189 m.