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Secure https://www.math rTest.aspx?testId= 1 671 37264&centerwin; =yes SSarkar: MATH 1113 Fall 2017 Wed 12 pm CRN 84115-051&93101-0 Quiz: Quiz 11 (Sec 5.4) Time Remaining: 00:13:18 S This Question: 1 pt 4-3 of 5 (0 complete) This Quiz: 5 A ship leaves its port and sails on a bearing of N34°30E, at speed 22.9 mph Another ship leaves the same port at the same time and sails on a bearing of S55°30'E, at speed 15.7 mph. 34° 30 5530 How far apart are the ships after 4 hours? miles (Round to the nearest integer as needed)

Explanation / Answer

distance travelled by ship 1 in 4 hours = 22.9*4 = 91.6 miles

distance travelled by ship 2 in 4 hours = 15.7*4 = 62.8 miles

the angle opposite to c is 180 - ( 34.30' + 55.30' )

= 90 degrees

now applying law of cosines

c^2 = a^2 + b^2 - 2ab cos C

c^2 = 91.6^2 + 62.8^2 - 2(91.6)(62.8) cos 90

c = 111

therefore, they are 111 miles apart

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