An airplane flies 200 km due west from city A to city B and then 240 km in the d

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Explanation / Answer

28.5 degrees N of W = 180-28.5 = 151.5 degrees CCW West = 180 degrees CCW Sum the x-components of both vectors: Dx = 200* Cos(180)+ 240* Cos(151.5)=-410.916107038872 Km ==>Cos(180)=-1 Sum the y-components of both vectors: Dy = 200*Sin(180)+ 240* Sin(151.5)=114.518102462303 Km ==>Sin(180)=0 Total distance from City A to City C is about: D = Sqrt(Dx^2+Dy^2) = 426.58 Km Direction = Arctan(Dy/Dx) = 164.42739185791600 degrees CCW Or, Relative to city A, city C is about 15.57 degrees North of West. Another way: cos(28.5 * pi / 180) = 0.878817113 sin(28.5 * pi / 180) = 0.47715876 Dx = -200 + (-240*0.878817113) = -410.91610712 km Dy = 0 + 240*0.47715876 = 114.5181024j km D = Sqrt(Dx^2+Dy^2) = 426.58 km Direction = Arctan(Dy/|Dx|)*180/pi = 15.57 degrees North of West

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