Найдите sin[latex]\alpha[/latex], если [latex]ctg\alpha=\frac{3}{4} ;180а меньше \alpha меньше 270а[/latex]

[latex]1+ctg^2\alpha=\frac1{\sin^2\alpha}\\ 1+\left(\frac34\right)^2=\frac1{\sin^2\alpha}\\ 1+\frac9{16}=\frac1{\sin^2\alpha}\\ \frac{25}{16}=\frac1{\sin^2\alpha}\\ {\sin^2\alpha}=\frac{16}{25}\\ \sin\alpha=\pm\frac45\\ 180^o<\alpha<270^o\Rightarrow\sin\alpha=-\frac45[/latex]