# Difference between revisions of "2019 AMC 12A Problems/Problem 19"

## Problem

In $\triangle ABC$ with integer side lengths, $$\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.$$ What is the least possible perimeter for $\triangle ABC$?

$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$

## Solution

We intend to use law of sines, so let's flip all the cosines (Sine is positive for $0\le x \le 180$, so we're good there).

$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=-\frac{\sqrt{15}}{4}$

These are in the ratio $2:3:4$, our minimal triangle has side lengths $2$, $3$, and $4$. $\boxed{(A) 9}$ is our answer.

-Rowechen Zhong