Our 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!
An airplane has an airspeed of 450 km/hr bearing N45E. The wind velocity is 30 km/hr in the direction N30W. What is the ground speed and what is its direction?.
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).
1 answer
1 answer