I have a mathematician problem where, I knew the 3 sides of a triangle, with these sides I can figer out what type of type of triangle is. What I realy want to find is the height of the triangle and another one "side".
Let me explain what I want, with the above picture. I knew a,b and c and I can calculate the 2 angles (the angle oposite of C, and the angle oposite of b, those on the black dots). I want to find out the red dots (the bottom red dot is is the other "side" I note before, from black dot to red dot distance).
So I want to to know the (x,y) for the red dot at the top, via the combination of height length, b OR c length (which also needs the two angles) in case to find the absolute right (x,y) for the red dot at the top. Thanks.
Use Heron's Formula for the area $A$ of a triangle
$$A=\sqrt{p\left( p-a\right) \left( p-b\right) \left( p-c\right) }=\frac{a\cdot h}{2},\tag{1}$$
where $h$ is the altitude length and $$2p=a+b+c\tag{2}$$ is the perimeter. A geometric proof of $(1)$ can be found in this post, in Portuguese. Solving $(1)$ for $h$ we find that
\begin{eqnarray*} h &=&\frac{2\sqrt{p\left( p-a\right) \left( p-b\right) \left( p-c\right) }}{a } \\ &=&\frac{2}{a}\sqrt{\frac{a+b+c}{2}\left( \frac{a+b+c}{2}-a\right) \left( \frac{a+b+c}{2}-b\right) \left( \frac{a+b+c}{2}-c\right) } \\ &=&\frac{1}{2a}\sqrt{\left( a+b+c\right) \left( b+c-a\right) \left( a+c-b\right) \left( a+b-c\right) }\tag{3} \end{eqnarray*}
Concerning coordinates, if the base $a$ is horizontal and the coordinates of the left vertex are $(x_{B},y_{B})=(0,0)$, then the coordinates of the upper vertex are
$$(x_{A},y_{A})=(\sqrt{c^{2}-h^{2}},h).\tag{4}$$
The coordinates of the foot of the altitude $h$ are thus $$(x_{N},y_{N})=(\sqrt{c^{2}-h^{2}},0).\tag{5}$$