The triangle $a, b, c$ is similar to the triangle $b, c, d$, where $a < b, b < c, c < d$.
I need to find a formula to find the possible length(s) of $d$ given $a, b, c$.
I've tried using:
$q=\cos^{-1}\left(\frac{c^{2}+a^{2}-b^{2}}{2ca}\right)$
$d_{1}=\sqrt{b^{2}+c^{2}-2bc\cos\left(q\right)}$
and
$r=\cos^{-1}\left(\frac{a^{2}+b^{2}-c^{2}}{2ab}\right)$
$d_{2}=\sqrt{b^{2}+c^{2}-2bc\cos\left(r\right)}$
but the angles never seem to match up
You want $~\displaystyle \frac{a}{b} = \frac{b}{c} = \frac{c}{d},~$ since the ratios of corresponding sides of similar triangles is constant, and since $~a < b < c < d.$
Therefore $~\displaystyle c = \frac{b^2}{a} \implies ~d = \frac{c^2}{b} = \frac{b^4}{a^2} \times \frac{1}{b} = \frac{b^3}{a^2}.~$
You also have the constraint that in any triangle, the larger side must be less than the sum of the shorter sides. So, you must have $~c < a+b, ~d < b+c.$
With free variables $~a,b~$ and constrained variables $~c,d$ you have that
$\displaystyle \frac{b^2}{a} < a + b \implies b^2 < a^2 + ab = a(a+b).$
$\displaystyle \frac{b^3}{a^2} < \frac{b^2}{a} + b \implies b^3 < ab^2 + a^2b = ab(a+b).$
Here, the common idea of a geometric sequence, $~1, x, x^2, x^3,~$ makes finding a satisfying set of values easy.
For example, taking $~x = 1.1,~$ leads to :