Odpowiedź i szczegółowe wyjaśnienie:
1090.
Korzystamy z "jedynki trygonometrycznej":
[tex]sin\alpha=\dfrac{\sqrt2}{3}\\\\sin^2\alpha+cos^2\alpha=1\\\\cos^2\alpha=1-sin^2\alpha\\\\cos^2\alpha=1-(\dfrac{\sqrt2}{3})^2\\\\cos^2\alpha=1-\dfrac{2}{9}\\\\\\cos^2\alpha=\dfrac79\\\\cos\alpha=\sqrt{\dfrac79}=\dfrac{\sqrt7}{\sqrt9}=\dfrac{\sqrt7}{3}[/tex]
1092.
[tex]cos\beta=30\% sin\beta\\\\tg\beta=\dfrac{sin\beta}{cos\beta}\\\\\\tg\beta=\dfrac{sin\beta}{\frac{3}{10}sin\beta}\\\\\\tg\beta=\dfrac{1}{\frac3{10}}=\dfrac{10}{3}=3\dfrac13[/tex]
