2019 AMC 12A Problem 19

Each sine is $\sqrt{1 - \cos^2}:$ $\sin A = \dfrac{3\sqrt{15}}{16},$ $\sin B = \dfrac{2\sqrt{15}}{16},$ $\sin C = \dfrac{4\sqrt{15}}{16}.$ By the Law of Sines the sides are in ratio $3 : 2 : 4.$ The smallest integer sides are $3, 2, 4,$ which satisfy the triangle inequality. The least perimeter is $3 + 2 + 4 = 9.$ Thus, the correct answer is A.