Answer:
The point C is 12.68 km away from the point A on a bearing of S23.23°W.
Step-by-step explanation:
Given that AB is 50 km and BC is 40 km as shown in the figure.
From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|
=|50 sin 80° - 40 cos 20°|=11.65 km
The length of y-component of AC = |AB cos 80° - BC sin 20°|
=|50 cos 80° - 40 sin 20°|= 5 km
tan[tex]\theta[/tex]= 5/11.65
[tex]\theta[/tex]=23.23°
AC= [tex]\sqrt{5^2+11.65^2}=12.68[/tex] km
Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.
