What is the proper heading for the plane in the Classic Airplane Problem?

[AxeluteZero]

Homework Statement

A small plane departs from point A heading for an airport 490 km due north at point B. The airspeed of the plane is 210 km/h and there is a steady wind of 50 km/h blowing directly toward the southeast.
(a) Determine the proper heading for the plane.
° west of north

(b) How long will the flight take?
h

Homework Equations

Law of Sines/Cosines?

v_f = v_i + a*t

The Attempt at a Solution

So I drew a picture.

You need a velocity vector. This vector must be the sum of the velocity vectors of the wind and the plane, which should also be drawn on the sketch. So I drew a simple sketch showing the plane's northward trek, an angle \Theta, another vector showing the wind and the resulting vector. So:

The first vector points due North (up), from point A.
The wind vector points toward point A, at an angle \Theta.
The resultant vector points from point B to the non-pointed end of the wind vector.

...so where in the world does the trig come in? I just don't know where to start plugging in numbers, since this isn't a right triangle it isn't a simple sin/cos problem with the Pythag. Theorem, right?

 

[rl.bhat]

In this problem, AB is the resultant velocity (Vr) of wind velocity (Vw)and proper velocity of the plane (Vp)
To make the problem easy, consider wind direction as x-axis.
Angle between wind direction and the resultant is ( 90 + 45) degrees.
Let θ be the angle between Vp and Vw. Then if you draw the vectors, you can see that,
Vr*sin(135) = Vp*sinθ ...(1)
Vr*cos(135) = Vp*cosθ - Vw ...(2)
Now solve for Vr and θ.

 

[AxeluteZero]

Ok I solved for Vr and for theta, but I can't use either of them since there's still the other variable in the equation. IE theta is still in the Vr equation (and I don't know it) and I have Vr in the theta equation. And plugging either equal to each other makes it unsolvable. I end up with a cos (-) plus sin (-).

Vr = Vr

\frac{210 * sin \Theta}{sin (135)} = \frac{210 (cos\Theta) - 50}{cos 135}

Help?

 

[rl.bhat]

Divide eq. 1 by 2 . You get
(Vp*sinθ) = Vp*cosθ - Vw
Square both the sides. You get
Vp^2*sin^2(θ) =( Vp*cosθ - Vw)^2
Vp^2*[1- cos^2(θ)] = ( Vp*cosθ - Vw)^2
Now simplify and solve the quadratic to find cosθ. Then find Vr.