Стороны треугольника равны 5,6,8. Найдите косинусы углов треугольника

Решите задачу:

${a}^{2} = {b}^{2} + {c}^{2} - 2bc \times cosА \\ cosА = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \\ cosА = \frac{ {8}^{2} + {5}^{2} - {6}^{2} }{2 \times 8 \times 5} = \\ = \frac{64 + 25 - 36}{80} = \frac{53}{80} \\ cosА = \frac{53}{80}$
$cosВ = \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac} \\ cosВ = \frac{ {5}^{2} + {6}^{2} - {8}^{2} }{2 \times 5 \times 6} = \\ = \frac{25 + 36 - 64}{60} = - \frac{3}{60} = - \frac{1}{20} \\ cosВ = - \frac{1}{20}$
$cosС = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} \\ cosС = \frac{ {6}^{2} + {8}^{2} - {5}^{2} }{2 \times 6 \times 8} = \\ = \frac{36 + 64 - 25}{96} = \frac{75}{96} = \frac{25}{32} \\ cosС = \frac{25}{32}$