[tex]\text{x oznaczamy nieznany bok trojkata}\\
\\
a) \\
2^2+x^2=(2\sqrt5)^2\\
4+x^2=20 /-4\\
x^2=16\\
x=4\\
\\
sin\alpha=\frac{4}{2\sqrt5}*\frac{\sqrt5}{\sqrt5}=\frac{4\sqrt5}{10}=\frac{2\sqrt5}5\\
cos\alpha=\frac{2}{2\sqrt5}=\frac{1}{\sqrt5}=\frac{\sqrt5}5\\
tg\alpha=\frac{4}2=2\\
ctg\alpha=\frac24=\frac12[/tex]
[tex]b)\\
(\sqrt2)^2+x^2=(\sqrt3)^2\\
2+x^2=3 /-2\\
x^2=1\\
x=1\\
\\
sin\alpha=\frac{1}{\sqrt3}=\frac{\sqrt3}3\\
cos\alpha=\frac{\sqrt2}{\sqrt3}=\frac{\sqrt6}3\\
tg\alpha=\frac{1}{\sqrt2}=\frac{\sqrt2}2\\
ctg\alpha=\frac{\sqrt2}1=\sqrt2[/tex]
[tex]c)\\
2^2+x^2=(2\sqrt{10})^2\\
4+x^2=40 /-4\\
x^2=36 \\
x=6\\
\\
sin\alpha=\frac{2}{2\sqrt{10}}=\frac{2\sqrt{10}}{20}=\frac{\sqrt{10}}{10}\\
cos\alpha=\frac{6}{2\sqrt{10}}=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}\\
tg\alpha=\frac{2}{6}=\frac13\\
ctg\alpha=\frac{6}2=3[/tex]
[tex]d)\\
x^2+1^2=(\sqrt5)^2\\
x^2+1=5 /-1\\
x^2=4\\
x=2\\
\\
sin\alpha=\frac{1}{\sqrt5}=\frac{\sqrt5}5\\
cos\alpha=\frac{2}{\sqrt5}=\frac{2\sqrt5}5\\
tg\alpha=\frac{1}2\\
ctg\alpha=\frac21=2[/tex]
