Find third triangle vertex given other 2 and lengths, without trigonometry
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Given the coorinates of points A, B and lengths of all sides, point C should be found. I have a solution which relies on tangent equation and cosine rule $φ_1 = arctan2(B_y - A_y, B_x - A_x)$ $φ_2 = arccosleft(dfrac{l_1^2 + l_3^2 - l_2^2}{2cdot l_1cdot l_3}right) $ $C = A + l_1cdot[cos(φ_1±φ_2)$ ; $sin(φ_1±φ_2)]$ Where $A_x$ , $A_y$ , $B_x$ , $B_y$ are the given coordinates and $l_1$ , $l_2$ , $l_3$ are lengths of $AC$ , $BC$ and $AB$ respectively. The $±$ sign is present because two triangles can satisfy the problem. Now, the issue with such solution is that this is a part of a bigger problem. I will have to substitute these equations into other equations which I will substitute into other equations... And in the end I'll have to find the invert function of the whole thing. Such soluti