What should I do further?

Question

I came across simple question, The length of all sides of a $\triangle{ABC}$ are in integral units. If length of $AB=10$ and $AC= 15$ then the number of distinct possible values of $BC$ is finite. We can simply apply Triangle Inequality, I tried to show creativity as follows:

Taking $\angle A= \theta$, and apply cosine rule. $BC^2=10^2 + 15^2 -300\cos\theta$

$BC=5\sqrt{13-12\cos\theta}$, here $\theta$ can't be greater than $180$, and its range is between $[-1,1]$.

Can I solve it further, using these? And what can be other interesting approach to this problem.

Answer 1

I think the Triangle Inequality is the cleanest method, applied twice to find the (excluded) lower bound of $5$ and the (excluded) upper bound of $25$, giving the full set of solutions for the length of $BC$ as $$\{6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24\}$$

A more creative method would of course be interesting, providing it didn't just boil down to find the two bounds above in a more obscure fashion.

Answer 2

$5\sqrt{13-12\cos\theta}=n\in\mathbb{N}$

$$12\cos\theta=13-{n^2\over25}$$ The absolute value of RHS is less than $12$ so $n=6,7,...,24$ are the only choices.