第14章 14.5

(1)

$\cos A = \frac{-a^2+b^2+c^2}{2bc} = \frac{-4+9+16}{24} = \frac{21}{24} = \frac{7}{8}$

$\cos B = \frac{a^2-b^2+c^2}{2ac} = \frac{4-9+16}{16} = \frac{11}{16}$

$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{4+9-16}{12} = -\frac{3}{12} = -\frac{1}{4}$

$\cos A : \cos B : \cos C = \frac{7}{8} : \frac{11}{16} : -\frac{1}{4} = 14 : 11 : -4$