в) cos A, tg А, если sin A=\frac{5}{13}.

в) Дано: $$sin A = \frac{5}{13}$$

Найти: $$cos A, tg A$$

Основное тригонометрическое тождество: $$sin^2 A + cos^2 A = 1$$

$$cos^2 A = 1 - sin^2 A = 1 - (\frac{5}{13})^2 = 1 - \frac{25}{169} = \frac{169-25}{169} = \frac{144}{169}$$

$$cos A = \sqrt{\frac{144}{169}} = \frac{12}{13}$$

$$tg A = \frac{sin A}{cos A} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$$

Ответ: $$cos A = \frac{12}{13}, tg A = \frac{5}{12}$$