Question #202825

A ship, Albatross, leaves a port on a bearing of N520E traveling at a speed of 42 km/h.  A second ship, Bountiful, leaves the same port at the same time on a bearing of N240W traveling at a speed of 29 km/h.  After 3 hours, how far apart are the two ships?


1
Expert's answer
2021-06-06T15:40:39-0400

Given



Both ships starts at the same time from port

to find: distance between both ships A and B after 3 hours.

Ship A moves in 3 hours =42×3=126km=42\times 3=126 km

Ship B moves in 3 hours =29×3=87km=29\times 3=87km

Angle between A and B is <AOB=52°+24°=76°=52 °+24 °= 76 °

By the law of cosine

AB2=OA2+OB2(OA)(OB)Cos(M<AOB)=1262+8722(126)(87)COS(76°)=18141.1043AB^{2}=OA^{2}+OB^{2}-(OA)(OB)Cos(M<AOB)\\=126^{2}+87^{2}-2(126)(87)COS(76 °)\\=18141.1043

AB=134.69KM

Distance between ship A and B after 3 hours is 134.69km


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