Question

Решите пожалуйста срочно!!!!

Answer 1

$tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}\\\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}=\frac43\\\\\frac{\sin^2\alpha}{1-\sin^2\alpha}=\frac{16}9\\9\sin^2\alpha=16-16\sin^2\alpha\\25\sin^2\alpha=16\\\sin^2\alpha=\frac{16}{25}\\\\\sin\alpha=\pm\frac45\\0<\alpha<\frac\pi2\Rightarrow\sin\alpha>0\\\sin\alpha=\frac45$