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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!

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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: 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:… in order to solve this problem:… – 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 $s=y'-n\cos A$.

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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
A is given in Degrees, and the function seems to work with degrees rather than radians. But the other half of the calculation works in radians, hence the question (I sorta understand the meaning/difference, but not usage, obviously). But... I am not quite getting the results I'm expecting from this formula/function. I'll post an example this evening (AEST) that shows what I'm seeing. I'm confident this function is correct, but some of my inputs may be incorrect or I'm making a false assumption. – Campbeln May 11 '12 at 4:22
@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

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