z twierdzenia Pitagorasa
a^2+b^2=c^2
a=\sqrt{c^2-b^2}
b=\sqrt{c^2-a^2}
I
a)
b=4 , c=7
a=\sqrt{7^2-4^2}=\sqrt{49-36}=\sqrt13
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sin\alpha=\frac{a}{c}=\frac{\sqrt{13}}{7}
cos\alpha=\frac{b}{c}=\frac{4}{7}
tg\alpha=\frac{a}{b}=\frac{\sqrt{13}}{4}
ctg\alpha=\frac{1}{\tg\alpha}=\frac{4}{\sqrt{13}}=\frac{4*\sqrt{13}}{\sqrt{13}*\sqrt{13}}=\frac{4\sqrt{13}}{13} odwrotność tg\alpha (1)
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sin\beta=\frac{b}{c}=\frac{4}{7}
cos\beta=\frac{a}{c}=\frac{\sqrt{13}}{7}
tg\beta=\frac{b}{a}=\frac{4}{\sqrt{13}}=\frac{4\sqrt{13}}{13} patrz (1)
ctg\beta=\frac{1}{tg\beta}=\frac{\sqrt{13}}{4}
b)
b=3 , c=4
a=\sqrt{4^2-3^2}=\sqrt{16-9}=\sqrt7
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sin\alpha=\frac{\sqrt7}{4}
cos\alpha=\frac{3}{4}
tg\alpha=\frac{\sqrt7}{3}
ctg\alpha=\frac{3}{\sqrt7}=\frac{3*\sqrt7}{\sqrt7*\sqrt7}=\frac{3\sqrt7}{7} (2)
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sin\beta=\frac{3}{4}
cos\beta=\frac{\sqrt7}{4}
tg\beta=\frac{3}{\sqrt7}=\frac{3\sqrt7}{7} patrz (2)
ctg\beta=\frac{\sqrt7}{3}
c)
b=4 , c=6
a=\sqrt{6^2-4^2}=\sqrt{36-16}=\sqrt{20}=\sqrt{4*5}=2\sqrt5
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sin\alpha=\frac{2\sqrt5}{4}=\frac{\sqrt5}{2}
cos\alpha=\frac{\not2^1\sqrt5}{\not6^3}=\frac{\sqrt5}{3}
tg\alpha=\frac{\not2^1\sqrt5}{\not4^2}=\frac{\sqrt5}{4}
ctg\alpha=\frac{4}{\sqrt5}=\frac{4*\sqrt5}{\sqrt5*\sqrt5}=\frac{4\sqrt5}{5}
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sin\beta=\frac{4}{6}=\frac{2}{3}
cos\beta=\frac{\not2^1\sqrt5}{\not6^6}=\frac{\sqrt5}{3}
tg\beta=\frac{\not4^2}{\not2^1\sqrt2}=\frac{2}{\sqrt2}=\frac{2*\sqrt2}{\sqrt2*\sqrt2}=\frac{2\sqrt2}{2}=\sqrt2
ctg\beta=\frac{1}{\sqrt2}=\frac{1*\sqrt2}{\sqrt2*\sqrt2}=\frac{\sqrt2}{2}
d)
a=5 , c=\sqrt{41}
b=\sqrt{(\sqrt{41})^2-5^2}=\sqrt{41-25}=\sqrt{16}=4
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sin\alpha=\frac{5}{\sqrt{41}}=\frac{5*\sqrt{41}}{\sqrt{41}*\sqrt{41}}=\frac{5\sqrt{41}}{41}
cos\alpha=\frac{4}{\sqrt{41}}=\frac{4\sqrt{41}}{41}
tg\alpha=\frac{5}{4}
ctg\alpha=\frac{4}{5}
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sin\alpha=\frac{4}{\sqrt{41}}=\frac{4\sqrt{41}}{41}
cos\beta=\frac{5}{\sqrt{41}}=\frac{5\sqrt{41}}{41}
tg\beta=\frac{4}{5}
ctg\beta=\frac{5}{4}