Zadanie 1
[tex]A=(-7,1)\\\\B=(1,-1)\\\\C=(-2,4)[/tex]
Długości boków:
[tex]|AB|=\sqrt{(-7-1)^2+(1-(-1))^2}=\sqrt{8^2+2^2}=\sqrt{64+4}=\sqrt{68}=2\sqrt{17}\\\\|AC|=\sqrt{(-7-(-2))^2+(1-4)^2}=\sqrt{5^2+3^2}=\sqrt{25+9}=\sqrt{34}\\\\|BC|=\sqrt{(1-(-2))^2+(-1-4)^2}=\sqrt{3^2+5^2}=\sqrt{9+25}=\sqrt{34}[/tex]
Mamy:
[tex](\sqrt{34})^2+(\sqrt{34})^2=(2\sqrt{17})^2[/tex]
Zatem trójkąt jest prostokątny i równoramienny. Miary kątów to:
[tex]|\sphericalangle CAB|=45^{\circ}\\\\|\sphericalangle CBA|=45^{\circ}\\\\|\sphericalangle ACB|=90^{\circ}[/tex]
Zadanie 2
[tex]A=(-4,-2\sqrt{3})\\\\B=(2,-2\sqrt{3})\\\\C=(-4,4\sqrt{3})[/tex]
Długości boków:
[tex]|AB|=\sqrt{(-4-2)^2+(-2\sqrt{3}-(-2\sqrt{3}))^2}=\sqrt{6^2+0^2}=6\\\\|AC|=\sqrt{(-4-(-4))^2+(-2\sqrt{3}-4\sqrt{3})^2}=\sqrt{0^2+(6\sqrt{3})^2}=6\sqrt{3}\\\\|BC|=\sqrt{(2-(-4))^2+(-2\sqrt{3}-4\sqrt{3})^2}=\sqrt{6^2+(6\sqrt{3})^2}=\sqrt{36+108}=\\\\=\sqrt{144}=\sqrt{12^2}=12[/tex]
Punkty A i C leżą na prostej równoległej do osi OY, natomiast punkty A i B leżą na prostej równoległej do osi OX. Zatem trójkąt jest prostokątny:
[tex]|\sphericalangle CAB|=90^{\circ}\\\\\text{tg }\sphericalangle ACB=\dfrac{|AB|}{|AC|}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\\\\|\sphericalangle ACB|=30^{\circ}\\\\|\sphericalangle ABC|=90^{\circ}-30^{\circ}=60^{\circ}[/tex]
Zadanie 3
[tex]A=(\sqrt{3},\sqrt{3})\\\\B=(3,\sqrt{3})\\\\C=(3+\sqrt{3},3+\sqrt{3})[/tex]
Długości boków:
[tex]|AB|=\sqrt{(\sqrt{3}-3)^2+(\sqrt{3}-\sqrt{3})^2}=\sqrt{(3-\sqrt{3})^2+0^2}=3-\sqrt{3}\\\\|AC|=\sqrt{(\sqrt{3}-(3+\sqrt{3}))^2+(\sqrt{3}-(3+\sqrt{3}))^2}=\\\\=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}\\\\|BC|=\sqrt{(3-(3+\sqrt{3}))^2+(\sqrt{3}-(3+\sqrt{3}))^2}=\\\\=\sqrt{(\sqrt{3})^2+3^2}=\sqrt{12}=2\sqrt{3}[/tex]
Liczymy miarę kąta ABC:
[tex]|AC|^2=|AB|^2+|BC|^2-2\cdot|AB|\cdot|BC|\cdot \cos\alpha\\\\18=(12-6\sqrt{3})+12-2\cdot2\sqrt{3}\cdot(3-\sqrt{3})\cdot\cos\alpha\\\\18=24-6\sqrt{3}-4\sqrt{3}\cdot(3-\sqrt{3})\cdot\cos\alpha\\\\(12-12\sqrt{3})\cdot\cos\alpha=6\sqrt{3}-6\\\\\cos\alpha=-\dfrac{1}{2}\\\\\alpha=120^{\circ}[/tex]
Liczymy miarę kąta CAB:
[tex]|BC|^2=|AB|^2+|AC|^2-2\cdot|AB|\cdot|AC|\cdot\cos\beta\\\\12=18+(12-6\sqrt{3})-2\cdot3\sqrt{2}\cdot(3-\sqrt{3})\cdot\cos\beta\\\\12=30-6\sqrt{3}-6\sqrt{2}\cdot(3-\sqrt{3})\cdot\cos\beta\\\\6\sqrt{2}\cdot(3-\sqrt{3})\cdot\cos\beta=18-6\sqrt{3}\\\\6\sqrt{2}\cdot\cos\beta=6\\\\\cos\beta=\dfrac{\sqrt{2}}{2}\\\\\beta=45^{\circ}[/tex]
Liczymy miarę kąta ACB:
[tex]\gamma=180^{\circ}-120^{\circ}-45^{\circ}=15^{\circ}[/tex]
