Triangle ABC's Angles

Let $t = 6 \sin \angle A = 3\sqrt{3} \sin \angle B = 2\sqrt{2} \sin \angle C$ , then $\sin \angle A = \frac{t}{6}$ , $\sin \angle B = \frac{t}{3\sqrt{3}}$ and $\sin \angle C = \frac{t}{2\sqrt{2}}$ . Thus, by the sine rule, we have

$BC:CA:AB = \sin \angle A: \sin \angle B: \sin \angle C = \frac{t}{6}: \frac{t}{3\sqrt{3}}: \frac{t}{2\sqrt{2}}\ = \sqrt{6}: 2\sqrt{2} : 3\sqrt{3}$

Let $BC = \sqrt{6}k$ , $CA = 2\sqrt{2}k$ and $AB = 3\sqrt{3}k$ , for some positive real number $k$ . By the cosine rule, we have

$\cos \angle A = \frac{(2\sqrt{2}k)^2 + (3\sqrt{3}k)^2 -(\sqrt{6}k)^2}{2\cdot 2\sqrt{2} k \cdot 3\sqrt{3} k} = \frac{29}{12\sqrt{6}}$

Thus $\sin ^2 \angle A = 1 - \cos ^2 \angle A = 1 - \frac{29^2}{12^2\cdot 6} = \frac{23}{864}$ . Hence $a + b = 23 + 864 = 887$ .