Question

3.Для острого угла а найдите ѕina, cosa, tgа, если ctg a=9/40.

Answer 1

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

$\displaystyle\bf\\tg\alpha =\frac{1}{Ctg\alpha } =1:\frac{9}{40} =1\cdot \frac{40}{9} =\frac{40}{9} =4\frac{4}{9}$