Physics-trig question velocity of a plane etc HELP

Question

A plane flies due north (90° from east) w/ a velocity of 110 km/h for 3 hours.A steady wind blows southeast at 35 km/h at an angle of 320 degrees from due east during this time. After 3 hours, where will the plane’s position be relative to its starting point?

I`m having trouble getting the answer I`ve started off with the formula v=sqrt{v(p)²+v(w)²-2v(p)v(w)cos α}
=sqrt(110²+35²-2*110*35cos45) = 88.8 km/h (rounded) Then did
s= vt=88.8*3=266.4 km
then tried getting the distance from the start by the following:
sqrt.(266.4^2 + 88. = 266.6 km N.E from starting point.
Although I have no idea why (trying to imitate a similar problem I did
this to get the angle from the start using tanL , L being the length
tanL = (266.6/88., which equals about degrees

Please explain . Agh thank you soo much

Answer 1

It is easy to use component vectors when solving problems of this type.

Velocity of plane = 0 i + 110 j

Velocity of wind = 35 cos 40 i - 35 sin 40 j = 26.81 i - 22.50 j

The resultant velocity is the sum of these two vectors = 26.81 i  + 87.5 j

The plane's position after 3 hours is 3 x (26.81 i + 87.5 j) =

Find the magnitude of this position vector by using the Pythagorean Theorem.

The direction of the vector is the Tan ^-1 (87.50/26.81).