Question

a girl starts from a point A and walks 285meter to B on a bearing of 078°.She then walks due south to a point C which is 307meter from A. What is the bearing of A from C, and what is |BC|?

Answer 1

In the triangle ABC, AB=285m, angle ABC=78º (because 078 bearing is 78 degrees from North making AB the same angle because BC is directed due South) and AC=307m.

We can use the sine rule to find angle ACB:

SinACB/285=sin78/307 so sinACB=285sin78/307=0.90805 approx, making ACB=65.24º approx. But ACB is the bearing of A from C=-65.24 or bearing 294.76 (just add 360 degrees).

Angle BAC=180-(78+65.24)=36.76º.

Using the sine rule again: BC/sin36.76=307/sin78, BC=307sin36.76/sin78=187.84m approx.