Odpowiedź :
Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]cos(\pi + \alpha) = \frac{5}{13} \\cos(\pi+\alpha) = - cos\alpha = \frac{5}{13}\\cos\alpha = -\frac{5}{13}\\[/tex]
[tex]sin(6\pi + \alpha) > 0 => sin\alpha > 0\\[/tex]
[tex]\sin\alpha > 0\\ \cos\alpha < 0\\=> tg\alpha < 0[/tex]
[tex]sin^2\alpha + cos^2\alpha = 1\\sin\alpha = \sqrt{1 - cos^2\alpha} = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}[/tex]
[tex]tg\alpha = \frac{sin\alpha}{cos\alpha} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5} =-2,4[/tex]