Question

7.

Annett, starting from a point A, walks 8km on bearing of 2650

to a point B. From B she walks on

bearing of 1350

for 9km to a point C. Find the bearing of A from C.

Answer 1

The bearing of a point is the number of degrees in the angle measured in a clockwise direction from the north line to the line joining the centre of the compass with the point. 

$$\overrightarrow{AB}=(8\sin(265\degree), 8\cos(265\degree))$$

$$\overrightarrow{BC}=(9\sin(135\degree), 9\cos(135\degree))$$

$$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$$

$$=(8\sin(265\degree)+9\sin(135\degree), 8\cos(265\degree)+9\cos(135\degree))$$

$$\tan \theta=\dfrac{8\cos(265\degree)+9\cos(135\degree)}{8\sin(265\degree)+9\sin(135\degree)}$$

The bearing of A from C

$$90\degree-\theta=90\degree-\tan^{-1}(\dfrac{8\cos(265\degree)+9\cos(135\degree)}{8\sin(265\degree)+9\sin(135\degree)})$$

$$\approx12.81\degree$$