A tetrahedron $ABCD$ has an equilateral base $ABC$ with coordinates $A = (0, 0, 0), B = (a, 0, 0), C = (\dfrac{a}{2},\dfrac{\sqrt{3} a}{2}, 0) $ , where the side length $a$ is given. The apex is point $D$,and we are given the angles $\angle ADB, \angle BDC, \angle CDA$. I want to determine the position of the apex $D$.
My approach:
Let $s_1 = DA, s_2 = DB, s_3 = DC $. And let $\theta_1 = \angle ADB, \theta_2 = \angle BDC, \theta_3 = \angle CDA $ , then using the law of cosines, we get
$ a^2 = s_1^2 + s_2^2 - 2 s_1 s_2 \cos \theta_1 $
$ a^2 = s_2^2 + s3^2 - 2 s_2 s_3 \cos \theta_2 $
$ a^2 = s_3^2 + s_1^2 - 2 s_3 s_1 \cos \theta_3$
We now have three quadratic equations in the three unknowns $s_1, s_2, s_3$. These can be solved numerically using, for example, Newton-Raphson multivariate method. Once these three lengths are determined, and they're all positive, then the apex can be found quite easily as follows. Let the apex $E = (x, y, z)$, then
$ s_1^2 = x^2 + y^2 + z^2 $
$ s_2^2 = (x - a)^2 + y^2 + z^2 $
$ s_3^2 = (x - \dfrac{a}{2})^2 + (y - \dfrac{\sqrt{3} a}{2})^2 + z^2 $
Then it follows by taking differences of the above equations, that
$ x = \dfrac{ a^2 + s_1^2 - s_2^2 }{2 a} $
$ y = \dfrac{ a^2 + s_1^2 - s_3^2 - a x } {\sqrt{3} a} $
And finally,
$ z = \sqrt{ s_1^2 - x^2 - y^2 } $
Whenever $s_1^2 - x^2 - y^2 \gt 0 $
From my numerical trials on this problems I found that, depending on the angles $\theta_1, \theta_2, \theta_3$, there could be no solutions, a single solution, or multiple solutions.
Any comments on this, or alternative solutions, are highly appreciated.