б) sin A, tg A, если cos A=\frac{\sqrt{13}}{5}.

б) Дано: $$cos A = \frac{\sqrt{13}}{5}$$

Найти: $$sin A, tg A$$

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

$$sin^2 A = 1 - cos^2 A = 1 - (\frac{\sqrt{13}}{5})^2 = 1 - \frac{13}{25} = \frac{25-13}{25} = \frac{12}{25}$$

$$sin A = \sqrt{\frac{12}{25}} = \frac{\sqrt{12}}{5} = \frac{2\sqrt{3}}{5}$$

$$tg A = \frac{sin A}{cos A} = \frac{\frac{2\sqrt{3}}{5}}{\frac{\sqrt{13}}{5}} = \frac{2\sqrt{3}}{\sqrt{13}} = \frac{2\sqrt{3}\sqrt{13}}{13} = \frac{2\sqrt{39}}{13}$$

Ответ: $$sin A = \frac{2\sqrt{3}}{5}, tg A = \frac{2\sqrt{39}}{13}$$