Question

Solve this A tourist starts off from town A and travels for 50km on a bearing of N80°W to town B.At town B ,he continues for another 40kmbon a bearing of N20°E toC. How far is C from A?

Answer 1

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$\theta$= 5/11.65

$\theta$=23.23°

AC= $\sqrt{5^2+11.65^2}=12.68$ km

Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.