Twierdzenie cosinusów Dla dowolnego trójkąta (oznaczenia jak na rysunku) prawdziwe są następujące zależności: $a^2=b^2+c^2-2bc\cos \alpha$   $b^2=a^2+c^2-2ac\cos \beta$   $c^2=a^2+b^2-2ab\cos \gamma$

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a) Mamy dane:

$a=5,b=8,\gamma ={60}^{\circ }$  

Z twierdzenia cosinusów:

$c^2=a^2+b^2-2ab\cos \gamma$  

$c^2=5^2+8^2-2\cdot 5\cdot 8\cdot \cos {60}^{\circ }$  

$c^2=25+64-80\cdot \frac{1}{2}$  

$c^2=89-40$  

$c^2=49$  

$c=7$  

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b) Mamy dane:

$a=5,b=3\sqrt{2},\gamma ={45}^{\circ }$  

Z twierdzenia cosinusów:

$c^2=a^2+b^2-2ab\cos \gamma$  

$c^2=5^2+{\left(3\sqrt{2}\right)}^2-2\cdot 5\cdot 3\sqrt{2}\cdot \cos {45}^{\circ }$  

$c^2=25+18-30\sqrt{2}\cdot \frac{\sqrt{2}}{2}$  

$c^2=43-30$  

$c^2=13$  

$c=\sqrt{13}$  

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c) Mamy dane:

$a=6,b=9,\gamma ={120}^{\circ }$  

Z twierdzenia cosinusów:

$c^2=a^2+b^2-2ab\cos \gamma$  

$c^2=6^2+9^2-2\cdot 6\cdot 9\cdot \cos {120}^{\circ }$  

$c^2=36+81-108\cdot \cos \left({180}^{\circ }-{60}^{\circ }\right)$  

$c^2=117-108\cdot \left(-\cos {60}^{\circ }\right)$  

$c^2=117-108\cdot \left(-\frac{1}{2}\right)$  

$c^2=117+54$  

$c^2=171$  

$c=3\sqrt{19}$  

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d) Mamy dane:

$a=2\sqrt{3},b=4,\gamma ={150}^{\circ }$  

Z twierdzenia cosinusów:

$c^2=a^2+b^2-2ab\cos \gamma$  

$c^2={\left(2\sqrt{3}\right)}^2+4^2-2\cdot 2\sqrt{3}\cdot 4\cdot \cos {150}^{\circ }$  

$c^2=12+16-16\sqrt{3}\cdot \cos \left({180}^{\circ }-{30}^{\circ }\right)$  

$c^2=28-16\sqrt{3}\cdot \left(-\cos {30}^{\circ }\right)$  

$c^2=28-16\sqrt{3}\cdot \left(-\frac{\sqrt{3}}{2}\right)$  

$c^2=28+24$  

$c^2=52$  

$c=2\sqrt{13}$