# Thread: Did I do this vector question right?

1. ## Did I do this vector question right?

A plane is heading S 70degrees W with a groundspeed of 625km/h. If the pilot is steering west at an airspeed of 665km/h, what must be the windspeed and wind direction?

This is the diagram I drew, forgive me if the scaling is off:

The magnitude of the wind I found was:

w^2 = a^2 + g^2 -2(a)(g)(cos110degrees)

Is this wrong? I want to know if I have the right answer for the magnitude before attempting to find the direction of the wind.

2. $w^2 = a^2 + g^2 - 2ag \cos(20)
$

3. Hello, kmjt!

Your diagram is incorrect . . .

A plane is heading S 70° W with a groundspeed of 625 km/h.
If the pilot is steering west at an airspeed of 665 km/h,
. . what must be the windspeed and wind direction?
Code:
                 665
A o - - - - - - - - - - - o O
\               20° *
\               *
\           *
\       *  625
\   *
o
B

The plane is flying from O to A at 665 km/hr.
. . $\angle AOB = 20^o$

The wind is blowing from A to B.

The plane's resultant flight is: . $OB = 625$ km/hr.

Law of Cosines: . $AB^2 \:=\:665^2 + 625^2 - 2(665)(625)\cos20^o \;=\;51,\!730.50897$

. . Hence: . $AB \;\approx\;227.4$

Then: . $\cos A \;=\;\frac{665^2 + 227.4^2 - 625^2}{2(665)(227.4)} \;=\;0.341588668$

. . Hnce: . $\angle A \;\approx\;70^o$

Therefore, the windspeed is 227.4 km/hr and its direction is: .S 20° E.

Interesting!
Triangle ABO is virtually a right triangle . . .
.

4. Doh thats what I got the first time I just thought it was wrong because usually we have to calculate something to find what angle the wind is blowing at.. Thanks