• Matéria: Matemática
  • Autor: mariah568
  • Perguntado 9 anos atrás

Seja um triangulo ABC, no qual a=5, b=v2e c= v17. calcule a medida do angulo Č.

Respostas

respondido por: JBRY
2
Boa Noite Maria!

Solução!

a^{2}=b^{2}+c^{2} -2.a.c.cos\alpha\\\\\\\ ( \sqrt{17})^{2} =5^{2}+ ( \sqrt{2})-2. \sqrt{2}.5 cos\alpha\\\\\\\ 17=25-10 \sqrt{2} cos\alpha\\\\\\\ 17-27=-10 \sqrt{2} cos\alpha\\\\\\\ -10=-10\sqrt{2} cos\alpha\\\\\\\  \dfrac{-10}{-10\sqrt{2} }=cos\alpha\\\\\\\  \dfrac{1}{\sqrt{2} }=cos\alpha


 \dfrac{1. \sqrt{2} }{ \sqrt{2}. \sqrt{2}  }= cos \alpha \\\\\\\
 \dfrac{1. \sqrt{2} }{ \sqrt{4}}= cos \alpha \\\\\\\
 \dfrac{\sqrt{2} }{ 2  }= cos \alpha \\\\\
\alpha=C=45\º


Boa noite!
Bons estudos!
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