Side Side Angle?

Simple, probably unoriginal problem, but I liked the idea.

First note that to maximize $\tan\angle A$ , we maximize $\angle A$ (it's true that when $\angle A > 90^{\circ}$ the value suddenly drops, but the situation doesn't let that happen, as we will see).

Draw segment $AB$ and consider the locus of all possible points $C$ . This is a circle with radius $15$ and center $B$ . Now it is clear that to maximize $\angle A$ , we just need to make $AC$ tangent to this circle, making $\angle ACB=90^{\circ}$

Now using Pythagorean Theorem we get $AC=\sqrt{AB^2-BC^2}=\sqrt{20^2+15^2}=5\sqrt{7}$

Thus $\tan\angle A = \dfrac{15}{5\sqrt{7}} = \dfrac{3}{\sqrt{7}}=\dfrac{3\sqrt{7}}{7}$ so our answer is $3+7+7=\boxed{17}$

By the Law of Sines we have:

$\frac{AB}{\sin \gamma} = \frac{BC}{\sin A} \Rightarrow \frac{20}{\sin \gamma} = \frac{15}{\sin A} \Rightarrow \sin A = \frac{3\sin \gamma}{4}$ and $\cos A = \frac{\sqrt{16 - 9\sin^{2} \gamma}}{4}$ .

We can express $\tan A$ as a function of $\gamma$ :

$tan A = f(\gamma) = \frac{3\sin \gamma}{\sqrt{16 - 9\sin^{2} \gamma}}$

and setting the first derivative equal to zero gives:

$f'(\gamma) = \frac{48\cos \gamma}{(16-9\sin^{2} \gamma)^{3/2}} = 0 \Rightarrow \gamma = \frac{(4n-1)\pi}{2}, \frac{(4n+1)\pi}{2}$ for $n \in \mathbb{Z}$

and checking the second derivative $f''(\gamma) = 24 \cdot \frac{9\sin 3\gamma - 5\sin \gamma}{(16-9\sin^{2} \gamma)^{5/2}}$ additionally yields:

$f''(\frac{(4n+1)\pi}{2}) < 0$ and $f''(\frac{(4n-1)\pi}{2}) > 0$

with the maximum value attained by the former. Hence, the maximum value for $\tan A$ is $f(\frac{(4n+1)\pi}{2}) =\frac{3(1)}{\sqrt{16-9(1^2)}} = \boxed{\frac{3\sqrt{7}}{7}}.$