Odpowiedź:
Szczegółowe wyjaśnienie:
Korzystamy ze wzorów:
M=(x,y) to promień wodzący punktu M obliczamy:
[tex]r=\sqrt{x^{2}+y^{2} } ,r > 0[/tex]
[tex]sin\alpha =\frac{y}{r}[/tex] [tex]cos\alpha =\frac{x}{r}[/tex] [tex]tg\alpha =\frac{y}{x}[/tex] [tex]ctg\alpha =\frac{x}{y}[/tex]
P=(√3,-1)
[tex]x=\sqrt{3}[/tex] [tex]y=-1[/tex]
[tex]r=\sqrt{(\sqrt{3} )^{2}+(-1)^{2} } =\sqrt{3+1} =\sqrt{4}=2[/tex]
[tex]sin\alpha =\frac{y}{r} =\frac{-1}{2}[/tex]
[tex]cos\alpha =\frac{x}{r} =\frac{\sqrt{3} }{2}[/tex]
[tex]tg\alpha =\frac{y}{x} =\frac{-1}{\sqrt{3} }=-\frac{\sqrt{3} }{3}[/tex]
[tex]ctg\alpha =\frac{x}{y} =\frac{\sqrt{3} }{-1} =-\sqrt{3}[/tex]
P=(-2,-6)
[tex]x=-2[/tex] [tex]y=-6[/tex]
[tex]r=\sqrt{x^{2} +y^{2} } =\sqrt{(-2)^{2} +(-6)^{2} } =\sqrt{4+36} =\sqrt{40} =\sqrt{4*10} =2\sqrt{10}[/tex]
[tex]sin\alpha =\frac{y}{r} =\frac{-6}{2\sqrt{10} }=-\frac{3}{\sqrt{10} }= -\frac{3}{\sqrt{10} }*\frac{\sqrt{10} }{\sqrt{10} } =-\frac{3\sqrt{10} }{10}[/tex]
[tex]cos\alpha =\frac{x}{r} =\frac{-2}{2\sqrt{10} } =-\frac{1}{\sqrt{10} } =-\frac{1}{\sqrt{10} }*\frac{\sqrt{10} }{\sqrt{10} } =-\frac{\sqrt{10} }{10}[/tex]
[tex]tg\alpha =\frac{y}{x} =\frac{-6}{-2} =3[/tex]
[tex]ctg\alpha =\frac{x}{y} =\frac{-2}{-6} =\frac{1}{3}[/tex]
