з) sin A, tg A, если cos A=\frac{2\sqrt{6}}{5}.

з) Дано: $$cos A = \frac{2\sqrt{6}}{5}$$

Найти: $$sin A, tg A$$

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

$$sin^2 A = 1 - cos^2 A = 1 - (\frac{2\sqrt{6}}{5})^2 = 1 - \frac{4 \cdot 6}{25} = 1 - \frac{24}{25} = \frac{1}{25}$$

$$sin A = \sqrt{\frac{1}{25}} = \frac{1}{5}$$

$$tg A = \frac{sin A}{cos A} = \frac{\frac{1}{5}}{\frac{2\sqrt{6}}{5}} = \frac{1}{2\sqrt{6}} = \frac{\sqrt{6}}{12}$$

Ответ: $$sin A = \frac{1}{5}, tg A = \frac{\sqrt{6}}{12}$$