A person walks 25 degrees North of East for 3.10km. How far due north and how fa

Question

A person walks 25 degrees North of East for 3.10km. How far due north and how far due east would she have to walk to arrive at the same location?

Explanation / Answer

(r, theta) --> (3.10 km, 25 degrees) [r = 3.10 km; theta = 25 degrees] x = rcostheta y = rsintheta x, in this case, would be the horizontal, or eastward distance, so the appropriate values would be substituted in the x equation to solve for it. x = 3.10 * (cos25) = 2.81 km y would be the vertical, or northward distance, so the appropriate values would be substituted in the y equation to solve for it. y = 3.10 * (sin25) = 1.31 km To verify the solutions, here is the optional step: (just to be sure!) r = sqrt (x^2 + y^2) r = sqrt (2.81^2 + 1.31^2) r = 3.0999 --> 3.10 km Therefore, the person must walk 1.31 km due north and 2.81 km due east to arrive at the same location.

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