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 Oct 11 comment Pythagorean theorem: $A^2 + A^2 = C^2$ How to solve for $A$? Correct, thank you for sharing! And thank you for catching the missed square in the comment above! Oct 11 comment Pythagorean theorem: $A^2 + A^2 = C^2$ How to solve for $A$? Perfect! Thank you for working it out fully! A ~= 19.79898987... for the lazy ;) May 11 comment Finding a point of an isosceles triangle *OR* Find the coordinates of the start-point of an angled line 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. May 11 comment Finding a point of an isosceles triangle *OR* Find the coordinates of the start-point of an angled line Excellent, thanks I'll give this a shot! One question: in this solution, is A (angle) in degrees or radians? May 10 comment Finding a point of an isosceles triangle *OR* Find the coordinates of the start-point of an angled line @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/… May 10 comment Finding a point of an isosceles triangle *OR* Find the coordinates of the start-point of an angled line 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).