tgα>0 czyli α ∈(0°, 90°) lub α∈(180°, 270°)
[tex]\text{tg\,}\alpha=\dfrac{\sin\alpha}{\cos\alpha}\\\\\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\sqrt3}4\qquad/\cdot\cos\alpha\\\\ \sin\alpha=\frac{\sqrt3}4\cos\alpha\\\\\\\sin^2\alpha+\cos^2\alpha=1\\\\(\frac{\sqrt3}4\cos\alpha)^2+\cos^2\alpha=1 \\\\ \frac3{16}\cos^2\alpha+\cos^2\alpha=1 \\\\ \frac{19}{16}\cos^2\alpha=1\qquad/\cdot\frac{16}{19} \\\\ \cos^2\alpha=\frac{16}{19}\\\\ \cos\alpha=\frac{4\sqrt{19}}{19}\quad\vee\quad\cos\alpha=-\frac{4\sqrt{19}}{19}[/tex]
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\ \sin^2\alpha+\frac{16}{19}=1\\\\ \sin^2\alpha=\frac{3}{19}\\\\ \sin\alpha=\frac{\sqrt3}{\sqrt{19}}\qquad\vee\quad \sin\alpha=-\frac{\sqrt3}{\sqrt{19}}\\\\ \sin\alpha=\frac{\sqrt{57}}{19}\qquad\vee\quad \sin\alpha=-\frac{\sqrt{57}}{19}\\\\\\ \text{ctg\,}\alpha=\dfrac1{\text{tg\,}\alpha}=\dfrac1{\frac{\sqrt3}4}=\frac4{\sqrt3}=\frac{4\sqrt3}3[/tex]
Jeśli α∈(0°, 90°) to [tex]\sin\alpha=\frac{\sqrt{57}}{19}\,,\ \ \cos\alpha=\frac{4\sqrt{19}}{19}\,,\ \ \text{ctg\,}\alpha=\frac{4\sqrt3}3[/tex]
Jeśli α∈(180°, 270°) to [tex]\sin\alpha=-\frac{\sqrt{57}}{19}\,,\ \ \cos\alpha=-\frac{4\sqrt{19}}{19}\,,\ \ \text{ctg\,}\alpha=\frac{4\sqrt3}3[/tex]
