she walks on a heading, not a bearing
(3,0)+(2.35,4.41) = (5.35,4.41)
so, the distance AC is √(5.35^2+4.41^2) = 6.93
bearing to C is E39.5°N
or, 50.5°
A student starts from A and walks 3km east to B, she then walk 5km on a bearing 152 degree from 13 . She reaches a point c . Find the distance and bearing of C from A.
2 answers
A Boy