# Vector Airplane Question

## Homework Statement

An airplane flies at 600km/h on a bearing of W38N while a wind of 75km/h from W25S is present.
Find the planes resultant velocity and direction

## The Attempt at a Solution

So heres what I did:

The Airplane: x = 600cos142 = -472.8
y = 600sin142 = 369.39

The Wind: x = 75cos25 = 67.973
y = 75sin25 = 31.69

Sum of x: -472.8 + 67.973 = -404.827
Sum of y: 360.30 + 31.69 = 401.09

Resultant would be using pythagorean theorem:

this gives me 569.877

and the angle would be theta = tan1- (401/-494.877) = -44.734 degrees

I think what I did is right (please correct if im wrong) but I was wondering about my angle. The angle is not below the horizontal right? I just dont understand why its negative. If i were to change it to bearing form would i do......... 90-44.734 = 45.26 degrees therefore on a bearing of N45.26W? Please clarify the solution I have found

I didn't check your numbers, but the method is correct. The angle is negative because the arctangent returns a value between $-\pi/2$ and $\pi/2$. You have to look at the signs of x and y to find the actual angle.