Niech dany będzie kąt wypukły 𝛼 w położeniu standardowym. Na drugim ramieniu kąta wybieramy punkt P(x,y) różny od punktu O(0,0). Wówczas:

$\text{tg}\alpha =\frac{y}{x},x\ne 0$  

$\text{ctg}\alpha =\frac{x}{y},y\ne 0$

$\sin \alpha =\frac{y}{\sqrt{x^2+y^2}}$  

$\cos \alpha =\frac{x}{\sqrt{x^2+y^2}}$  

---

$\text{a)}A\left(12,5\right)\Rightarrow x=12,y=5,\sqrt{x^2+y^2}=\sqrt{{12}^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$  

$\sin \alpha =\frac{5}{13}$  

$\cos \alpha =\frac{12}{13}$  

$\text{tg}\alpha =\frac{5}{12}$  

$\text{ctg}\alpha =\frac{12}{5}$  

---

$\text{b)}B\left(-2,2\sqrt{3}\right)\Rightarrow x=-2,y=2\sqrt{3},\sqrt{x^2+y^2}=\sqrt{{\left(-2\right)}^2+{\left(2\sqrt{3}\right)}^2}=\sqrt{4+12}=\sqrt{16}=4$  

$\sin \alpha =\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$  

$\cos \alpha =\frac{-2}{4}=-\frac{1}{2}$  

$\text{tg}\alpha =\frac{2\sqrt{3}}{-2}=-\sqrt{3}$  

$\text{ctg}\alpha =\frac{-2}{2\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$  

---

$\text{c)}C\left(-\sqrt{3},-\sqrt{6}\right)\Rightarrow x=-\sqrt{3},y=-\sqrt{6},\sqrt{x^2+y^2}=\sqrt{{\left(-\sqrt{3}\right)}^2+{\left(-\sqrt{6}\right)}^2}=\sqrt{3+6}=\sqrt{9}=3$  

$\sin \alpha =-\frac{\sqrt{6}}{3}$  

$\cos \alpha =-\frac{\sqrt{3}}{3}$  

$\text{tg}\alpha =\frac{-\sqrt{6}}{-\sqrt{3}}=\sqrt{2}$  

$\text{ctg}\alpha =\frac{-\sqrt{3}}{-\sqrt{6}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$  

---

$\text{d)}D\left(15,-8\right)\Rightarrow x=15,y=-8,\sqrt{x^2+y^2}=\sqrt{{15}^2+{\left(-8\right)}^2}=\sqrt{225+64}=\sqrt{289}=17$  

$\sin \alpha =-\frac{8}{17}$  

$\cos \alpha =\frac{15}{17}$  

$\text{tg}\alpha =-\frac{8}{15}$  

$\text{ctg}\alpha =-\frac{15}{8}$