Question

3.Для острого угла α найдите sin α, cos α, tg α, если ctg α =9/40.​

Answer 1

$Ctg\alpha=\frac{9}{40}\\\\tg\alpha=\frac{1}{Ctg\alpha}=\frac{40}{9}=4\frac{4}{9}\\\\1+Ctg^{2}\alpha=\frac{1}{Sin^{2}\alpha} \\\\Sin^{2}\alpha=\frac{1}{1+Ctg^{2}\alpha} =\frac{1}{1+(\frac{9}{40})^{2}} =\frac{1}{1+\frac{81}{1600}} =\frac{1}{\frac{1681}{1600}}=\frac{1600}{1681}\\\\Sin\alpha=\sqrt{\frac{1600}{1681}}=\frac{40}{41} \\\\Cos\alpha=\sqrt{1-Sin^{2}\alpha}=\sqrt{1-(\frac{40}{41})^{2}}=\sqrt{1-\frac{1600}{1681}}=\sqrt{\frac{81}{1681}}=\frac{9}{41}$

$Otvet:\boxed{Sin\alpha=\frac{40}{41};Cos\alpha =\frac{9}{41};tg\alpha=4\frac{4}{9}}$