Question

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

Answer 1

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