[tex]tg\alpha = \frac{11}{12}\\sin\alpha = \frac{5}{6}\\tg\alpha = \frac{sin\alpha}{cos\alpha}\\\frac{11}{12} = \frac{\frac{5}{6}}{cos\alpha}\\11cos\alpha = \frac{5}{6} * 12\\11cos\alpha=10\\cos\alpha = \frac{10}{11}\\sin^2\alpha + cos^2\alpha = 1\\(\frac{5}{6})^2+(\frac{10}{11}) ^2 = 1\\\frac{25}{36} + \frac{100}{121} = 1\\\frac{3025}{4356} + \frac{3600}{4356} = 1\\\frac{6625}{4356} \neq 1\\[/tex]
Odp. Nie istnieje taki kąt ostry.
