Question

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

Answer 1

Boa Noite Maria!

Solução!

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


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


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Bons estudos!