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March 28, 2015

March 28, 2015

Posted by **VEJ** on Wednesday, July 13, 2011 at 1:25pm.

I have gotten this far:

A= 450cos(45), 450sin(45) = (318.20, 318.20)

W= 30cos(120), 30sin(120) = (-15, 25.98)

A+W= 303.2 +344.18

|A+W|= 458.7 km.hr

tan-1(344.18/303.2)= 48.62

How do I change 48.62 to degrees east of north, which is the actual direction if the aircraft relative to due north (round to the nearest tenth degree).

- trig -
**Henry**, Thursday, July 14, 2011 at 6:52pmOur answers are identical:

458.7km/hr @ 48.62 deg. CCW.

90 - 48.62 = 41.38 deg,E of N.

You did a good job on this problem!

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