Answer to Question #142871 in Trigonometry for Raffy

Question #142871
A ship sails on course N30oE finds a lighthouse bearing 120o true, 5
miles away.


1.) How far should the ship sails to find the lighthouse due south?
1
Expert's answer
2020-11-09T20:54:26-0500


A= "x"

C = 30"\\degree"

B= 120"\\degree"

A = 180"\\degree" - 30"\\degree" - 120"\\degree"

A = 30"\\degree"

"\\frac{sinA}{5miles}" = "\\frac{sinB}{b}"


"\\frac{sin30}{5miles}" = "\\frac{sin120}{b}"


"\\frac{1\/2}{5miles}" = "\\frac{\\sqrt3\/2}{b}"


1/2 ( b )= "\\sqrt3\/2" (5miles)

b = 5 "\\sqrt3" miles

the ship sails to find the lighthouse due south is 5"\\sqrt3" miles


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