a girl walks from a point A to a point B,20Km Away From A In Northeast Direction And WalkS To a point C on the bearing of 125degree,she is 80km away from the starting point,find her bearing from the starting point

Answer 1

Northeast ----> 45°

After you make your sketch you will have triangle ABC, where
Angle B = 100°, AB = 20 and AC = 89
you have to find angle A

By the sine law, first find angle C
SinC/20 = sin100/80
Angle C = ...

angle A = 180 - 100 - C
From there use your diagram to find the "bearing"

Answer 2

Given: AB = 20km[45o],
BC = BC[125o],
AC = 80km[Xo],

Draw triangle ABC:
B = 125 - 45 = 80o,
C = 125 - 90 = 35o,
A = 180 - (80+35) = 65o.

Law of sine:
BC/sin65 = 80/sin80,
BC = 74 km.

AC = AB + BC = 20km[45o] + 74km[125o],
AC = (20*sin45+74*sin125) + (20*Cos45+74*Cos125)I = 74.8 - 28.3i,
AC = 80km[-69.3o] = 80km[69.3o] E. of S. = 80km[110.7o] CW(bearing).