# Finding a point of an isosceles triangle *OR* Find the coordinates of the start-point of an angled line

How do I find the coordinates ?/? (green star) given n, A (angle) and x'/y' (red circle)? NOTE: The n on the left side is vertical, while the n on the right side is at A angle from this vertical line.

I'm sure if I knew the correct term to ask for I'd be able to locate the answer within the site, but alas, I was stretching to use the term "isosceles triangle" correctly =) Many thanks for any help you can provide!

-
So, you have a point at distance $n$ from a known point, and you want to work out where that unknown point is. Think about it a minute and you'll see that's impossible. But, wait - what's the other side of the angle? Is that other line of length $n$ supposed to be vertical? That would do the trick, give you enough information to answer the question. –  Gerry Myerson May 10 '12 at 7:28
This looks to be a potential solution, but it looks like it solves for the other end of the line. With this problem, it will always be the "start-point" to solve for. See: forums.codeguru.com/showthread.php?t=472141 sin(theta) = (y2 - y1) / L so: y2 = [L * sin(theta)] + y1 and cos(theta) = (x2 - x1) / L so: x2 = [L * cos(theta)] + x1 (L = Length of line, (x1, y1) = start point, (x2, y2) = end point & theta = angle). –  Campbeln May 10 '12 at 7:30
@Gerry Myerson: Yes, the "left" n is indeed vertical! I'm trying to solve this problem: stackoverflow.com/questions/10508022/… in order to solve this problem: stackoverflow.com/questions/10392658/… –  Campbeln May 10 '12 at 7:33

Let the unknown point have co-ordinates $(r,s)$. Then $\sin A=(x'-r)/n$ and $\cos A=(y'-s)/n$, where $(x',y')$ is the known point. So $r=x'-n\sin A$ and $s=y'-n\cos A$.
Excellent, thanks I'll give this a shot! One question: in this solution, is A (angle) in degrees or radians? –  Campbeln May 11 '12 at 1:28
You wrote that $A$ was given. Was it given to you in degrees, or in radians? –  Gerry Myerson May 11 '12 at 3:52
@Ross, my formulas are not in any particular units. If $A$ is given as 30 degrees, then the formula calls for the sine of 30 degrees. If $A$ is given as pi-over-6 radians, then the formula calls for the sine of pi-over-6 radians. An angle is what it is, no matter how you measure it, and its sine is opposite over hypotenuse, no matter how you measure it. –  Gerry Myerson May 11 '12 at 6:23