Question

Для острого угла найдите sin a, cos a, tg a если ctg a = 2/5

Answer 1

$sinx=\frac{1}{\sqrt{ctg^2x+1}}=\frac{1}{\sqrt{(\frac{2}{5})^2+1}}=\frac{5}{\sqrt{29}}\\cosx=\frac{ctgx}{\sqrt{ctg^2x+1}}=\frac{\frac{2}{5}}{\sqrt{(\frac{2}{5})^2+1}}=\frac{2}{\sqrt{29}}\\tgx=\frac{1}{ctgx}=\frac{1}{\frac{2}{5}}=2.5$