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A ship leaves a port and travels 21km on a bearing of 32 and then 45km on a bearing of 287 calculate its distance from the port and the bearing of the port from the ship

User Nandithakw
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1 Answer

4 votes

Answer:

44.5 km on a bearing of 134.1°

Explanation:

A suitable vector calculator will tell you the location of the ship from port is ...

21∠32° +45∠287° = 44.5∠-45.9°

The bearing of the port is in the reverse direction, so 180° added to this:

180° +(-45.9°) = 134.1°

The distance from port is 44.5 km. The bearing of the port is 134.1°.

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Additional comment

The distance from port is perhaps most easily found using the Law of Cosines. It is the third side of a triangle with sides 21 and 45. The included angle is (287° -(180° +32°)) = 75°. The distance is then ...

c = √(a² +b² -2ab·cos(C)) = √(21² +45² -2·21·45·cos(75°)) ≈ √1976.83

≈ 44.5 km

The angle between the 32° bearing and the bearing from the port to the ship can be found using the Law of Sines.

B = arcsin(b/c·sin(C)) = arcsin(45/44.5sin(75°)) ≈ 77.9°

so, the bearing angle from the ship to the port is ...

180° +(32° -77.9°) = 134.1°

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Bearings are measured clockwise from north. A diagram can be helpful.

User Ersan
by
8.7k points
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