How do you solve this vector word problem?

Question

An airplane ends up 540 miles and 20 degrees north of east from its departure point. If there was a steady wind of 30 mph from the northwest during the entire flight, then find the magnitude and direction that the plane would have gone if there had been no wind.

Answer 1

The airplane as a vector would be --> a =〈540cos(20),540sin(20)〉

The wind as a vector would be --> w =〈30cos(135),30sin(135)〉

The resultant vector would be --> a+w =〈540cos(20)+30cos(135), 540sin(20)+30sin(135)〉

Finding the magntiude would be ||a+w|| = √[[540cos(20)+30cos(135)]²+[540cos(20)+30cos(135)]²] ≈ 528.02

The direction would be θ = tan^-1(w/a) = tan^-1([540sin(20)+30sin(135)]/[540cos(20)+30cos(135)]) ≈ 22.95º

Hope this helped!

Answer 2

can say

rcos(θ)+(30t)cos(pi/4)=540cos(pi/9)

rsin(θ)-(30t)sin(pi/4)=540sin(pi/9)

r is distance and θ heading with east as zero

of plane without wind. pi/9<θ<10pi/9

r(cos(θ)+sin(θ))=540(cos(pi/9)+sin(pi/9))

if θ=pi/2,,,,r=540(cos(pi/9)+sin(pi/9))

if θ=pi/4,,,,r=(270√2)(cos(pi/9)+sin(pi/9))

another way to look at problem is to fill in how much time the trip took (missing in description)

If trip is specified as being one hour, then fly plane at 528 mph at 22.9 degrees to reach 540 miles at 22 degrees when wind is 30 mph at -45 degrees., if trip is specified as being 10 hours, then fly plane at 49.4 mph at 53.3 degrees to reach 540 miles at 22 degrees with 30 mph wind at -45 degrees. In the ten hour case, if there was no wind, plane after ten hours would be 494 miles at 53.3 degrees.

note that in original answer provided,,,, r(cos(θ)+sin(θ))=540(cos(pi/9)+sin(pi/9)),,,,if 53.3 degrees is value of theta, then r=494. check.Now it is learned that the condition for r=494, theta=53.3 degrees is trip takes ten hours.