Printable View
I am working with line plotting in a form.
I need to find an angle between two lines AB,BC joined at B.
AB and BC lengths are not same.
I know (x,y) values of A,B and C.
How to find angle at B?
and also
I have Trignometric "TanA" value.
How to find angle 'A' ?
in VB code point, can some body help me?
sqr((Ax-Bx)^2 + (Ay-By)^2) = length AB
sqr((Cx-Bx)^2 + (Cy-By)^2) = length BC
TanA - isn't that the Arc Tan of an angle ?? Say B? ;)
Or do you mean Tan(A) = degrees/radians?
I'm trying to remember back around ten years ;)
Post up again later :)
Vince
The sandpaper library will have a derived math class in it. It will contain functions for all the derived math equations detailed in MSDN.
It has already been coded. Alpha Launch date: 01 Dec 04.
I have a line with known x,y values. that is x1,y1 and x2,y2.
I need to shift that line with same slope and lenght to new point to x3,y3.
that is new line coordinates shall be x3,y3 and x4,y4.
how do I do it in VB point of view. any help pl.
Very Urgent.
x3 - x1 = diffx
y3 - y1 = diffy
x4 = x2 + diffx
y4 = y2 + diffy