Answer to Question #202825 in Trigonometry for cole

Question

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?

Answer 1

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 $=42\times 3=126 km$

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

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

By the law of cosine

$AB^{2}=OA^{2}+OB^{2}-(OA)(OB)Cos(M<AOB)\\=126^{2}+87^{2}-2(126)(87)COS(76 °)\\=18141.1043$

AB=134.69KM

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