Question

Oblicz wartości funkcji trygonometrycznych katów ostrych trójkąta prostokątnego o bokach długości:
a) 9, 40, 41
b) 2, 4, 2√5
c) √13, 2√3, 5

Answer 1

Odpowiedź:Szczegółowe wyjaśnienie:[tex]sin(\alpha) = \frac{a}{c}\\cos(\alpha) =\frac{b}{c}\\tg(\alpha) = \frac{a}{b}\\ctg(\alpha)=\frac{b}{a}\\\\sin(\beta) = cos(\alpha) = \frac{b}{c}\\cos(\beta) = sin(\alpha) = \frac{a}{c}\\tg(\beta) = ctg(\alpha) = \frac{b}{a}\\ctg(\beta)=tg(\alpha) = \frac{b}{a}\\\\[/tex]Tak więca)  [tex]sin(\alpha) = cos(\beta) = \frac{9}{41} \\cos(\alpha) = sin(\beta) = \frac{40}{41}\\tg(\alpha) = ctg(\alpha) = \frac{9}{40}\\ctg(\alpha} = tg(\beta) = \frac{40}{9}\\[/tex]b)[tex]sin(\alpha) = cos(\beta) = \frac{2}{2\sqrt{5} }=\frac{1}{\sqrt{5}} \\cos(\alpha) = sin(\beta) = \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}}\\tg(\alpha) = ctg(\alpha) = \frac{2}{4} = \frac{1}{2}\\ctg(\alpha} = tg(\beta) = \frac{4}{2} = 2[/tex]c)[tex]sin(\alpha) = cos(\beta) = \frac{\sqrt{13}}{5} \\cos(\alpha) = sin(\beta) = \frac{2\sqrt{3}}{5}\\tg(\alpha) = ctg(\alpha) = \frac{\sqrt{13}}{2\sqrt{3}} = \sqrt{\frac{13}{12}}\\ctg(\alpha) = tg(\beta) = \frac{2\sqrt{3}}{\sqrt{13}} = \sqrt{\frac{12}{13}}\\[/tex]