\sin(90-\alpha) =\frac{3}{10}
\sin(90-\alpha) =\cos\alpha wzory redukcyjne
czyli
\cos\alpha=\frac{3}{10}
\sin^2\alpha+\cos^2\alpha=1
\sin^2\alpha=1-\cos^2\alpha=1-(\frac{3}{10})^2=1-\frac{9}{100}=\frac{91}{100}
\sin\alpha=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}\approx0,9539
\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{0,9539}{0,3}\approx=3,1797
\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{3,1797}\approx0,3145
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