Thread: Rotation of a triangle

Hi,

I have a triangle with the points A( 5,-2) B(0,0) and C(4,2) . I have to rotate the triangle so that the line BA sits on the positive x axis. I am stuck, how to work out the values for sin, cos and tan?

I have worked out the lengths of those lines BA = sqroot[29], AC = sqroot[17] and BC = sqroot[20]. After that I have no idea where to start, any help is appreciated.

use the sine and cosine laws to find two angles, then draw line AB on the x-axis, since you know how long it is, and then lines AC and BC off of it having calculated the right angles

a^2 = b^2 + c^2 - 2bcCosA

a/SinA = b/SinB = c/SinC

Thanks, I think I get it. It says though I don't need the angle but I should express the rotation as an equation. Do you know about that?