Witaj :)
Wiemy, że;
[tex]\sin\alpha=\frac{2}{5}\ oraz\ \alpha\in(90^\circ;180^\circ)[/tex]
Kąt alfa należy do drugiej ćwiartki, zatem tylko sinus będzie dodatni.
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\(\frac{2}{5})^2+\cos^2\alpha=1\\\\\frac{4}{25}+\cos^2\alpha=1 \\\\\cos^2\alpha=1-\frac{4}{25} \\\\\cos^2\alpha=\frac{21}{25} \implies \cos\alpha=\frac{\sqrt{21} }{5}\ \vee cos\alpha =- \frac{\sqrt{21} }{5}\\\\Poniewaz\ \alpha\in (90^\circ;180^\circ) \implies \cos\alpha<0,\ wiec:\\\\\large \boxed{\cos\alpha =-\frac{\sqrt{21}}{5} }[/tex]
[tex]tg\alpha=\frac{\sin\alpha}{\cos\alpha}\\\\tg\alpha=\frac{\frac{2}{5} }{-\frac{\sqrt{21}}{5} } \\\\tg\alpha=\frac{2}{5}\cdot (-\frac{5}{\sqrt{21}} )\\\\tg\alpha=-\frac{2}{\sqrt{21}}\\\\\large \boxed{tg\alpha=-\frac{2\sqrt{21}}{21} }[/tex]