0° < α < 90°
a)
[tex]cos\alpha = \frac{5}{13}\\\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha = 1 - cos^{2}\alpha = 1 - (\frac{5}{13})^{2} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169}\\\\sin\alpha = \sqrt{\frac{144}{169}} = \frac{12}{13}\\\\\\tg\alpha = \frac{sin\alpha}{cos\alpha} =\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}\\\\\\ctg\alpha = \frac{1}{tg\alpha} = \frac{5}{12}[/tex]
b)
[tex]tg\alpha = \frac{4}{15}\\\\\\\frac{sin\alpha}{cos\alpha} = \frac{4}{5}\\\\5sin\alpha = 4 cos\alpha \ \ /:4\\\\cos\alpah = \frac{5}{4}sin\alpha\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha + (\frac{5}{4}sin\alpha)^{2} = 1\\\\sin^{2}\alpha + \frac{25}{16}sin^{2}\alpha = 1\\\\\frac{16}{16}sin^{2}\alpha + \frac{25}{16}sin^{2}\alpha = 1\\\\\frac{41}{16}sin^{2}\alpha = 1 \ \ /\cdot\frac{16}{41}\\\\sin^{2}\alpha = \frac{41}{16}[/tex]
[tex]sin\alpha = \sqrt{\frac{16}{41}} = \frac{4}{\sqrt{41}}\cdot\frac{\sqrt{41}}{\sqrt{41}} = \frac{4\sqrt{41}}{41}\\\\\\cos\alpha = \frac{5}{4} sin\alpha = \frac{5}{4}\cdot\frac{4\sqrt{41}}{41} = \frac{5\sqrt{41}}{41}\\\\ct\alpha = \frac{1}{tg\alpha} = \frac{5}{4}[/tex]
