Odpowiedź:
a)
[tex]sin \alpha = \frac{1}{5} \\ \\ {sin}^{2} \alpha + {cos}^{2} \alpha = 1 \\ ( \frac{1}{5} {)}^{2} + {cos}^{2} \alpha = 1 \\ \frac{1}{25} + {cos}^{2} \alpha = 1 \\ {cos}^{2} \alpha = \frac{24}{25} \\ cos \alpha = \sqrt{ \frac{24}{25} } = \frac{2 \sqrt{6} }{5} \\ \\ [/tex]
[tex]tg \alpha = \frac{sin \alpha }{cos \alpha } = \frac{ \frac{1}{5} }{ \frac{2 \sqrt{6} }{5} } = \frac{1}{5} \times \frac{5}{2 \sqrt{6} } = \frac{1}{2 \sqrt{6} } = \frac{ \sqrt{6} }{12} \\ \\ ctg \alpha = \frac{1}{tg \alpha } = \frac{1}{ \frac{ \sqrt{6} }{12} } = \frac{12}{ \sqrt{6} } = \frac{12 \sqrt{6} }{6} = 2 \sqrt{6} [/tex]
b)
[tex]cos \alpha = \frac{5}{13} \\ \\ {sin}^{2} \alpha + {cos}^{2} \alpha = 1 \\ {sin}^{2} \alpha + ( \frac{5}{13} {)}^{2} = 1 \\ {sin}^{2} \alpha + \frac{25}{169} = 1 \\ {sin}^{2} \alpha = \frac{144}{169} \\ sin \alpha = \sqrt{ \frac{144}{169} } = \frac{12}{13} [/tex]
[tex]tg \alpha = \frac{sin \alpha }{cos \alpha } = \frac{ \frac{12}{13} }{ \frac{5}{13} } = \frac{12}{13} \times \frac{13}{5} = \frac{12}{5} \\ \\ ctg \alpha = \frac{1}{tg \alpha } = \frac{1}{ \frac{12}{5} } = \frac{5}{12} [/tex]
c)
[tex]tg \alpha = \frac{3}{2} \\ \frac{sin \alpha }{cos \alpha } = \frac{3}{2} \\ sin \alpha = \frac{3}{2} cos \alpha \\ {sin}^{2} \alpha + {cos}^{2} \alpha = 1 \\ ( \frac{3}{2} cos \alpha {)}^{2} + {cos}^{2} \alpha = 1 \\ \frac{9}{4} {cos}^{2} \alpha + {cos}^{2} \alpha = 1 \\ \frac{13}{4} {cos}^{2} \alpha = 1 \\ {cos}^{2} \alpha = \frac{4}{13} \\ cos \alpha = \sqrt{ \frac{4}{13} } = \frac{2}{ \sqrt{13} } = \frac{2 \sqrt{13} }{13} [/tex]
[tex]sin \alpha = \frac{3}{2} \times \frac{2 \sqrt{13} }{13} = \frac{3 \sqrt{13} }{13} \\ \\ ctg \alpha = \frac{1}{tg \alpha } = \frac{1}{ \frac{3}{2} } = \frac{2}{3} [/tex]
