Question

Как решить?25 баллов

Answer 1

Ответ:

$\displaystyle sin\alpha =\frac{5\sqrt{34}}{34}=\frac{5}{\sqrt{34}}\\\\1+ctg^2\alpha =\frac{1}{sin^2\alpha }\ \ \ \Rightarrow \ \ \ \ ctg^2\alpha =\frac{1}{sin^2\alpha }-1\\\\\\ctg^2\alpha =\frac{1}{\dfrac{25}{34}}-1=\frac{34}{25}-1=\frac{9}{25}\\\\\\90^\circ <\alpha <180^\circ \ \ \ \ \Rightarrow \ \ \ ctg\alpha <0\ \ ,\ \ ctg\alpha =-\sqrt{\frac{9}{25}}=\boxed{-\frac{3}{5}\ }$