Introductory Vector Problem- Airplane emergency landing

[Jarvis88]

Homework Statement

[/B]
A plane leaves the airport in Galisteo and flies 180 km at 67.0 degrees east of north; then it changes direction to fly 255 km at 49.0 degrees south of east, after which it makes an immediate emergency landing in a pasture.

When the airport sends out a rescue crew, how far should this crew fly to go directly to this plane

The Attempt at a Solution

I assume I need the angle as measured counterclockwise from the x-axis,
90deg - 67deg =23deg

Ax = 180cos23 =165.69 km
Ay = 180sin23 =70.33 km

Bx = 255cos49 =167.295 km
By = -255sin49 = -192.45 (because I have my vector pointed in the negative y direction along the positive x-axis)

Ax+Bx = 332.99 km
Ay+By = -122.12 km

R= sqrt [(332.99^2)+(122.12^2)]
R= 310 km

I'm not sure where I've gone wrong. I also used Ax=180sin67 and Ay=180cos67, which give the same numbers (because they're complimentary angles?). I didn't round my numbers when I calculated everything; I put in the exact numbers when I added and took the square root.
Mastering Physics says I'm wrong, but doesn't provide a hint as to why.

 

[Orodruin]

R= sqrt [(332.99^2)+(122.12^2)]

This number must be at least 333 ...

 

[Jarvis88]

This number must be at least 333 ...

Thank you! I rechecked my math, and somehow I must not have entered it into my calculator correctly. This may be a silly question, but how would I know it's supposed to be at least 333? Is it because 333 was the largest squared number and it was added to the other number?