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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?

Sagot :

xero099

Answer:

Town C is approximately 58.356 kilometers from town A.

Step-by-step 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:

[tex]\vec r = (50\,km)\cdot (-\sin80^{\circ},\cos 80^{\circ})+(40\,km)\cdot (\sin 20^{\circ}, \cos 20^{\circ})[/tex]

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

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

[tex]\|\vec r\| = \sqrt{(-35.560\,km)^{2}+(46.270\,km)^{2}}[/tex]

[tex]\|\vec r\| \approx 58.356\,km[/tex]

Town C is approximately 58.356 kilometers from town A.

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