An airplane leaves an airport and flies to west 120 miles and then 150 miles in the direction S 44.1°W. How far is the plane from the airport (round to the nearest mile)?

Correct result:

d =  251 nm

#### Solution:

$x = 120+150 \cos(-44.1 ^\circ ) = 227.719 \ nm \ \\ y = 150 \sin(-44.1 ^\circ ) = -104.387 \ nm \ \\ \ \\ d^2 = x^2 + y^2 \ \\ d = \sqrt{ 227.719^2 + (-104.387)^2 } = 251 \ \text{nm}$

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