I am trying to determine the the coordinates of the end point of a line. This line is inclined by $\theta$ .
In this example we have $a = (4.5,4)$, $\theta = 45$ , $ab = 3$
I wanted to calculate $b = (?,?)$ , I tried calculating it with some trigonometry rules $b_x = \sin(\theta * AB) $ And $b_y = \cos(\theta * AB)$ but from the result I believe that it is wrong .
I also asked my math teacher, he gave me this equation with poor explanation that this is just some simple Analytic geometry $ \|\vec{a}\| = \sqrt{(x_b-ya)^2 + (y_b-y_a)^2} $
I don't know what this really mean, I think this calculate the distance of two points, and I think it's wrong two because it's not related to the angle $\theta$ .
Did you remember to shift the $(b_x,b_y)$ over a vector $(a_x,a_y)$? The actual formulae are
$$b_x-a_x=|ab|\cos\theta,\ b_y-a_y=|ab|\sin\theta.$$ Also note that sines are vertical, cosines are horizontal.
One final complication might be that your angle is measured from the negative $X$-axis. This does not influence the sine but it reverses the sign of the cosine.