SOLUTION: A girl starts from 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 |B

∠BAD = ∠EAD-∠EAB = 90°-78° = 12°

|BD|/|AB| = cos(∠BAD)

|BD| = |AB|∙sin(∠BAD) = 286∙sin(12°) = 59.46274357

|AD|² = |AB|²-|BD|² = 285² - 59.46274357² = 77689.18213

|DC|² = |AC|²-|AD|² = 307²-77689.18213 = 16559.81787



sin(∠DAC) = |DC|/|AC| = 128.6849559/307 = 0.4191692375

∠DAC = sin-1(0.4191692375) = 24.78214909°

Bearing from A = ∠EAC = ∠EAD+∠DAC = 90°+24.78214909° = 114.7821491°

|BC| = |BD|+|DC| = 59.46274357+128.6849559 = 188.1376995m

Edwin

I just realized I found the wrong bearing above.  You asked for the bearing of A
from C and I misread and found the bearing of C from A instead. What you want
is the outside (reflex) angle at C, indicated by the green arc below.

Since ∠DAC = 24.78214909°, ∠ACD = 90°-∠DAC = 90°-24.78214909°=65.21785091°, and
the green outside (reflex angle) is 360°-∠ACD = 360°-65.21785091° =
294.7821491° 



Edwin