Question

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 40km on a bearing of N20°E to C. How far is C from A?

Answer 1

Town C is approximately 58.356 kilometers from town A.

Explanation:

According to the statement, we understand that tourist travels from town A to town B (50 kilometers) at a direction of 80º west of north and from town B to town C (40 kilometers) at a direction of 20º east of north. Vectorially speaking, the resultant from town A to town C is described by the following formula:

`\vec r = (50\,km)\cdot (-\sin80^(\circ),\cos 80^(\circ))+(40\,km)\cdot (\sin 20^(\circ), \cos 20^(\circ))`

`\vec r = (-35.560,46.270)\,[km]`

The distance from town A to town C is the magnitude of vector reported above, which is now calculated by Pythagorean Theorem:

`\|\vec r\| = \sqrt{(-35.560\,km)^(2)+(46.270\,km)^(2)}`

`\|\vec r\| \approx 58.356\,km`

Town C is approximately 58.356 kilometers from town A.