Suppose you know the coordinates for points A and B of a triangle. We can refer to those coordinates as (Ay,Ax) and (By,Bx). Also, suppose you know the X coordinate for point C (Cx) but do not know the Y coordinate. Is there a way to find the Y coordinate? You do not know any information about the angles or the sides (besides side c which is the line between points A and B).
I came up with a solution, and I think it ought to work, but neither Wolfram-Alpha nor Symbolab can solve it. I don't know if it is because of how convoluted the equation ends up becoming or if it is because it is actually wrong.
The first step is to find the length of the sides. Using the distance formula, we can find the length of side c and write the length of sides a and b as equations in terms of Cy since we already know Cx (a=sqrt(Cx-Ax)^2+(Cy-Ay)^2), with x and y changing to Bx and By for b). Then, we can use the Law of Cosines to find angle C, also in terms of only Cy. The equation becomes C=arccos((c^2-b^2-a^2)/(-2*a*b)). At this point, having a and b written in terms of Cy predictably starts making it pretty messy. We also do the same thing with one other angle (Angle A for the purposes of demonstration), giving us A=arccos((a^2-b^2-c^2)/(-2*a*b)). Then, we can use the Law of Sines to relate everything to side c, which should allow us to solve for Cy since side c is a constant. SinA/a=SinC/c -> c=(aSinC)/SinA. Taking the Sin of either angle here means taking the Sin of an Arccos, which means we could rewrite the Sin's as Sqrt(1-(whatever we were taking the arccos of)^2), which gives us an equation we could theoretically simplify and solve, since the entire thing is only in terms of Cy now. However, this is all but impossible to solve by hand without making a mistake, and every online solver I've tried cannot do it. Is there something wrong with what I set up?
If anyone is curious why I am doing this, my marketing class is using the SABRE simulation. We are given four points on a graph showing us how much customers prefer our products (y axis) at a given attribute level (x axis; the attributes are things like price, power, speed, etc.). Pretty much, it tells us that at price x (or power x, etc.) the customers are y interested in our product. Sometimes, customers just don't care about a certain category (e.g. they're rich so the price isn't important) so the y value is really low, which means we don't want to waste resources making sure to give them what they want in that category. Using some unrelated data they gave us, I was able to figure out what the ideal level is for each attribute (I know exactly what price customers want, exactly how much power, etc.) but I still only have a vague estimate of how important each attribute is to them. The reason I want to find the Y coordinate is so that I can make my products in the most cost effective way possible and not waste resources focusing on unimportant attributes.
You can find what the numbers mean in this google doc, but here is an equation I set up using real numbers: 60.76005533=((((sqrt(((262.4-236.28)^2)+((x-0.14)^2))))Sin(((arccos(((((sqrt(((262.4-297.04)^2)+((x-0.058)^2))))^2)-(((sqrt(((262.4-236.28)^2)+((x-0.14)^2))))^2)-(((sqrt(((262.4-236.28)^2)+((x-0.14)^2))))^2))/(-2((sqrt(((262.4-236.28)^2)+((x-0.14)^2))))((sqrt(((262.4-236.28)^2)+((x-0.14)^2))))))))))/(Sin(((arccos(((((sqrt(((262.4-236.28)^2)+((x-0.14)^2))))^2)-(((sqrt(((262.4-236.28)^2)+((x-0.14)^2))))^2)-(((sqrt(((262.4-297.04)^2)+((x-0.058)^2))))^2))/(-2((sqrt(((262.4-236.28)^2)+((x-0.14)^2))))*((sqrt(((262.4-297.04)^2)+((x-0.058)^2)))))))))))
Best Answer
The triangle is not unique, for example $A=(0,0), B=(0,1), C=(3,y)$ always form a triangle regardless of the $y$.
The issue with your approach is, you'll always end up in getting an equation where each term involves $C_y$ cancel. That is simply because the validity of those identities does not depend on the value of $C_y$.
Let me use a simpler analogy of things similar to what you were trying to do.
Question: Can we find the value of $x$ if $x+y=0$?
Attempt: we know $x=-y$, we also know $2x=-2y$. Because $-2y=2\cdot(-y)$, plug it back we get $2x=2\cdot x$, oops it cancels.