[tex]Zad. 1\\\text{Przyprostokatne: } 2, \sqrt5\\\text{Przeciwprostokatna: } \\4+5=c^2\\c=3\\\\sin\alpha=\frac{2}3\\cos\alpha=\frac{\sqrt5}3\\tg\alpha=\frac{2}{\sqrt5}=\frac{2\sqrt5}5\\ctg\alpha=\frac{\sqrt5}2[/tex]
[tex]Zad. 2[/tex]
[tex](1dm)^2+(2dm)^2=d^2\\d^2=1dm^2+4dm^2\\d^2=5dm^2\\d=\sqrt5dm[/tex]
[tex]sin\alpha=\frac{1}{\sqrt5}=\frac{\sqrt5}5\\cos\alpha=\frac{2}{\sqrt5}=\frac{2\sqrt5}5\\tg\alpha=\frac{1}{2}\\ctg\alpha=\frac{2}1=2\\\\sin\beta=\frac{2}{\sqrt5}=\frac{2\sqrt5}5\\cos\beta=\frac{1}{\sqrt5}=\frac{\sqrt5}5\\tg\beta=\frac{2}1=2\\ctg\beta=\frac{1}2[/tex]
[tex]Zad. 3\\sin45=\frac{a}{3\sqrt6}\\\frac{\sqrt2}2=\frac{a}{3\sqrt6}\\2a=3\sqrt{12}\\2a=6\sqrt{3}/:2\\a=3\sqrt3\\\\sin60=\frac{3\sqrt3}y\\\frac{\sqrt3}2=\frac{3\sqrt3}y\\y=\frac{6\sqrt3}{\sqrt3}=6\\\\cos60=\frac{3\sqrt3-x}{y}\\\frac12=\frac{3\sqrt3-x}{6}\\2(3\sqrt3-x)=6\\6\sqrt3-2x=6 /-6\sqrt3\\-2x=6-6\sqrt2\\-2x=2(3-3\sqrt2) /:(-2)\\x=-(3-3\sqrt2)\\x=-3+3\sqrt2\\\\\bold{y=6}\\\bold{x=3\sqrt2-3}[/tex]
