15．在$\triangle ABC$中，内角$A$，$B$，$C$的对边分别为$a$，$b$，$c$，在$\triangle ABC$的周长为$7$，面积为$\frac{3\sqrt{7}}{4}$，且$ab + 8c = 28$，则$c =$．
解：依题$a + b + c = 7$，$s = \frac{1}{2}ab\sin C = \frac{3\sqrt{7}}{4}$
$\therefore c = 7 - a - b$，$ab\sin C = \frac{3\sqrt{7}}{2}$
$\therefore ab + 8c = 28$
$\therefore ab + 8(7 - a - b) = 28$
$\therefore a + b = \frac{28 + ab}{8}$
$\therefore \cos C = \frac{a^{2} + b^{2} - (7 - a - b)^{2}}{2ab} = \frac{14(a + b) - 2ab - 49}{2ab}$
$= \frac{\frac{7(28 + ab)}{4} - 2ab - 49}{2ab} = \frac{- \frac{ab}{4}}{2ab} = - \frac{1}{8}$
$\therefore \sin C = \frac{3\sqrt{7}}{8}$
$\therefore ab = 4$，$a + b = 4$，$\therefore c = 3$