### Instruction: In triangle $ABC,$ $AB = 20$ and $BC = 15.$  Find the largest possible value of $\tan A.$.### Response: I know that $\tan A$ is the ratio of the opposite side to the adjacent side in a right triangle, but this is not necessarily a right triangle.
However, I also know that $\tan A$ is positive when $A$ is acute and negative when $A$ is obtuse.
Therefore, the largest possible value of $\tan A$ occurs when $A$ is as close as possible to a right angle, without being equal to it.
That means $AC$ should be as close as possible to the hypotenuse of a right triangle with legs $AB$ and $BC,$ without being equal to it.
By the Pythagorean theorem, the hypotenuse of such a triangle would be $\sqrt{20^2 + 15^2} = \sqrt{625} = 25.$
So, $AC$ should be slightly less than $25,$ say $24.9999.$
Then, using the law of cosines, I can find $\cos A = \frac{20^2 + 24.9999^2 - 15^2}{2 \cdot 20 \cdot 24.9999} \approx 0.6.$
Since $\cos A$ is positive, I can use the positive square root to find $\sin A = \sqrt{1 - \cos^2 A} \approx 0.8.$
Finally, I can find $\tan A = \frac{\sin A}{\cos A} \approx 1.333.$
This is the largest possible value of $\tan A,$ because if I increase $AC$ any further, $A$ would become obtuse and $\tan A$ would decrease.