Skorzystamy z tożsamości trygonometrycznej: [tex]\text{tg\,}\alpha=\dfrac{\sin\alpha}{\cos\alpha}[/tex]
oraz jedynki trygonometrycznej: [tex]\sin^2\alpha+\cos^2\alpha=1[/tex]
[tex]\alpha\in(90^o,180^o)\quad\implies\quad\sin\alpha>0\ \ \wedge\ \ \cos\alpha<0[/tex]
[tex]\text{tg\,}\alpha=-\dfrac{20}{10,5}=-\dfrac{200}{105}=-\dfrac{40}{21}\\\\ -\dfrac{40}{21}=\dfrac{\sin\alpha}{\cos\alpha}\qquad\ /\cdot\cos\alpha\\\\ \sin\alpha=-\frac{40}{21}\cos\alpha\\\\\\ \sin^2\alpha+\cos^2\alpha=1\\\\ (-\frac{40}{21}\cos\alpha)^2+\cos^2\alpha=1\\\\ \frac{1600}{441}\cos^2\alpha+\cos^2\alpha=1\\\\ \frac{2041}{441}\cos^2\alpha=1\qquad\ /:\frac{2041}{441}\\\\\cos^2\alpha=\frac{441}{2041}\qquad \wedge\qquad \cos\alpha<0\\\\\cos\alpha=-\sqrt{\frac{441}{2041}}=-\frac{21\sqrt{2041}}{2041}[/tex]
[tex]\sin\alpha=-\frac{40}{21}\cdot(-\frac{21\sqrt{2041}}{2041})=\frac{40\sqrt{2041}}{2041}[/tex]
