Airplane Wind problem Airplane flying at 400 mph at an angle of 30 deg encounters a wind.  The resultant velocity of the airplane is 475.3 mph at an angle of 27.18 deg.  What was direction of the wind.
I set this up as tan (27.18) = (Ry/Rx)
But the total Ry contains the y vector of the plane and the y vector of the wind, and since I don't know either wind magnitude nor direction I end up with 2 unknowns in both Ry and Rx.
Then I thought about using law of cosines, but again, come up with one equation, 2 unknowns.
Any thoughts?
 A: It may be easier to treat the direction of the plane as along one axis, like the $x$ axis:
$$\vec{v_0} = 400 \hat{x}.$$
Then the new velocity is
$$\vec{v} = 475.3 (\cos (-2.82^{\circ}) \hat{x} + \sin (-2.82^{\circ}) \hat{y}) \approx 474.72 \hat{x} - 23.38 \hat{y}.$$
So the speed of the wind is
$$\sqrt{74.72^2 + (-23.38)^2} \approx 78.29 \text{ MPH}$$
and its direction is
$$\arctan(-23.38/74.72) + 30^{\circ} \approx -17.37^{\circ} + 30^{\circ} = 12.63^{\circ}.$$
A: A fairly standard method is to write all the vectors in their $x$ and $y$ components.
From the aircraft's speed and direction, $400$ mph at an angle of $30$ degrees,
you get components $A_x$ and $A_y.$
From the resulting velocity and direction, $475.3$ mph at an angle of $27.18$ degrees,
you get components $R_x$ and $R_y.$
This is all from standard formulas for taking $x$ and $y$ components of vectors, for example $A_x = 400 \cos(30^\circ),$ $A_y = 400 \sin(30^\circ).$
Now you know that original velocity + wind should give the resultant velocity,
that is, $A + W = R.$
So now you have
\begin{align}
A_x + W_x &= R_x,\\
A_y + W_y &= R_y.
\end{align}
Two equations, two unknowns ($W_x$ and $W_y$).
Finally, take the components $W_x$ and $W_y$ and turn them back into a speed and direction. Now is the time to bring out the tangent function.
Don't forget the Pythagorean Theorem for the length.

The Law of Cosines works too. It does not tell you the angle, but it does tell you the length of the unknown side of the vector triangle.
Once you have the length you can apply the Law of Sines with two known sides and one known angle to get one unknown angle of the triangle,
and from that you can determine the direction of the vector.
You do not always need to be able to solve for two things simultaneously when you have two unknowns!
If you can do one at a time, that's often enough.
