Zadanie 2 Oblicz miary kątów \alfa i \beta.
a)
a=6
b=3
z twierdzenia Pitagorasa:
c=\sqrt{a^2+b^2}=\sqrt{6^2+3^2}=\sqrt{45}=\sqrt{9*5}=3\sqrt5 przeciwprostokątna
sin\alpha=\frac{a}{c}=\frac{6}{3\sqrt5}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}
cos\alpha=\frac{b}{c}=\frac{3}{3\sqrt5}=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}
tg\alpha=\frac{a}{b}=\frac{6}{3}=2
ctg\alpha =\frac{1}{tg\alpha}=\frac{1}{2}
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sin\beta=\frac{b}{c}=\frac{3}{3\sqrt5}=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}
cos\beta=\frac{a}{c}=\frac{6}{3\sqrt5}=\frac{2\sqrt5}{5}
tg\beta=\frac{b}{a}=\frac{3}{6}=\frac{1}{2}
ctg\beta=\frac{1}{tg\beta}=\frac{1}{0,5}=2
b)
a=2 , c=5 , b=?
z twierdzenia Pitagorasa:
b=\sqrt{c^2-a^2}=\sqrt{5^2-2^2}=\sqrt{21}
sin\alpha=\frac{a}{c}=\frac{2}{5}
cos\alpha=\frac{b}{c}=\frac{\sqrt{21}}{5}
tg\alpha=\frac{a}{b}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}
ctg\alpha=\frac{b}{a}=\frac{21}{2}
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sin\beta=cos\alpha=\frac{\sqrt{21}}{5}
cos\beta=sin\alpha=\frac{\sqrt{21}}{5}
tg\beta=\frac{\sqrt{21}}{2}
ctg\beta=tg\alpha=\frac{2\sqrt{21}}{21}