I reply to Math4life post as a test of my own understanding. I am not a teacher nor am I experienced at explaining or solving maths problems. Below is my response anyway.
The wording of the maths problem was
An airliner is flying in the direction North 20 degrees East with an airspeed of 300mph. Its ground speed and true course are 350mph and 30 degrees respectively(RESULTANT). Use a vector analysis to determine the direction and speed of the wind. Include a diagram.
First, there is a need to change some terms used.
Instead of the term "airspeed " ......I will use the term
air velocity.
Instead of the term "ground speed and true course" I will use the term
ground velocity.
Instead of the term "wind speed and direction" ...I will use the term
wind velocity.
Math4life stated his problem as:
350mph/30 is the resultant/true course
I setup x components and y components to add below:
x:
300cos(70)+wind[cos(theta)]
y:
and 300sin(70)+wind[sin(theta)]
I don't know how to substitute properly to solve for the two variables theta and wind - which is the mph speed of the wind which is the second vector and not the resultant. Please help this is due tomorrow morning.
I made an attempt at getting a solution to this maths problem and it seems Math4life got stuck on formula transposition
(sorry, the proper name for the process escapes me at this point.)
Basically Math4life wrote Rx=Vx+Wx and Ry=Vy+Wy.
(R=ground velocity, V=planes air velocity, W=Wind Velocity, x = x-axis component of)
The unknown vector value W should be on its own. So, subtracting the
x-component or y-component of V from both sides of the equals sign
would be more useful. Math4life would be better off writing
Rx - Vx = Wx and Ry - Vy= Wy.
The angle of the wind velocity can be evaluated as
the reverse tan of (y-component / x-component).
Math4life would have given this angle the label 'theta'.
This gives tha angle of the wind vector from the the x-axis.
The format of the angle will need to be change to N/S degrees E/W form.
The x- and y- components can be put together and the vector magnitude
can be evaluated by W = \sqrt{(Wx^{2}+Wy^{2})}.
Please let me know of any errors and suggestions of better formating.
