I have a triangulation problem I'm trying to solve for a work project. It's been ages since I've taken a math class so assume that I've forgotten everything because I probably have.
So here's the situation:
I have a tetrahedron ($ABCD$). I know all three angles ($\angle ABC \angle ACB \angle BAC$) and the lengths of all three sides ($\overline{AB}$ $\overline{AC}$ $\overline{BC}$) of the base ($\triangle ABC$). I know all three angles around the peak ($D$) ($\angle ADB$ $\angle ADC$ $\angle BDC$). With that I can deduce the dihedral angles between three faces ($\triangle ABD$ to $\triangle ACD$, $\triangle ACD$ to $\triangle BCD$, $\triangle ABD$ to $\triangle BCD$).
How can I solve for the lengths of any of the sides between the base and the peak ($\overline{AD}$ $\overline{BD}$ $\overline{CD}$), or any of the remaining angles ($\angle ABD$ $\angle BAD$ $\angle ACD$ $\angle CAD$ $\angle BCD$ $\angle CBD$)?
It seems to me with a base having a fixed, known shape and size, and fixed angles around the fourth point, it should be possible to directly deduce the rest of the angles and side lengths.
Thanks in advance.
-Jon
UPDATE:
I feel like there's something lurking in the Law of Sines. Since I have one angle and its opposite side for faces $\triangle ABD$ $\triangle ACD$ and $\triangle BCD$, I know the ratio between the sines of the unknown angles and their opposite faces. Also, from the Law of Sines as applied to a tetrahedron I know the ratio between the sines of angles from adjacent faces opposite their shared side:
$$\frac{\sin(\angle ABD)}{\sin(\angle ACD)}=\frac{\sin(\angle ADB)\cdot\sin(\angle ABC)}{\sin(\angle ADC)\cdot\sin(\angle ACB)}$$
To satisfy the ratios within both $\triangle ABD$ and $\triangle ACD$, AND the ratio between $\sin(\angle ABD)$ and $sin(\angle ACD)$ should force only one possibility for $\overline{AD}$, right?
Best Answer
One way is to do it numerically for example with wolframalpha. use Li Li solution
$$\frac{AD^2+CD^2-AC^2}{2AD\cdot CD}=\cos(\angle ADC)$$
$$\frac{BD^2+CD^2-BC^2}{2BD\cdot CD}=\cos(\angle BDC)$$
$$\frac{AD^2+BD^2-AB^2}{2AD\cdot BD}=\cos(\angle ADB)$$
like here, it's a example of 2 second degree equations