An airplane flies 200 km due west from city A to city B and then 295 km in the d

Question

An airplane flies 200 km due west from city A to city B and then 295 km in the direction of 29.0° north of west from city B to city C.

(a) In straight-line distance, how far is city C from city A?

I got the correct answer for this, it was 479.82 km

(b) Relative to city A, in what direction is city C?

in degrees north of west

I keep doings this problem and getting part A correct but I can't seem to get B. I am taking the inverse COS of 200/479.82 and getting something around 65, i don't remember exactly, but it keeps telling me it's wrong.

Some help would be greatly appreciated!!! Thanks so much!

Oh! Also, why am I supposed to use COS in particular to find the angle of A as opposed to SIN or TAN?

Explanation / Answer

Let City A = origin (coordinates 0,0)
Let North-South be Y-axis (North is positive Y-direction)
Let East-West be X-axis (East is positive X-direction)
Displacements are relative to positive x-direction

X: (200 km)(cos 180) + (295 km)(cos 151) = -458.012 km
Y: (200 km)(sin 180) + (295 km)(sin 151) = 143.018 km

D^2 = X^2 + Y^2
D = 479.822 km

Bearing:
tan (theta) = Y/X
theta = arctan(143.018/ -458.012)
theta = -17.34 = 17.34* North of West

So:
D =479.822 km; Bearing =17.34* North of West

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