• Matéria: Matemática
  • Autor: toptoptop38
  • Perguntado 3 anos atrás

O valor de a na imagem abaixo, é:

Anexos:

Respostas

respondido por: albertrieben
2

Resposta:

teorema dos senos

a/sen(60) = 6/sen(45)

a*sen(45) = 6*sen(60)

a*√2/2 = 6√3/2

a = 6√3/√2 = 6√6/2 = 3√6 (B)


toptoptop38: Obrigado ^^
respondido por: CyberKirito
1

\Large\boxed{\begin{array}{l}\underline{\rm Lei~dos~Senos}\\\sf sendo~a,b~e~c~os~lados~de~um~tri\hat angulo~qualquer\\\sf~e~\hat A,\hat B~e~\hat C\\\sf os~\hat angulos~opostos~a~\boldsymbol{a,b}~\sf e\boldsymbol{~c}~respectivamente\\\sf ent\tilde ao\\\sf\dfrac{a}{sen\hat A}=\dfrac{b}{sen\hat B}=\dfrac{c}{sen \hat C}=2R\\\\\sf onde~2R~\acute e~o~di\hat ametro~da~circunfer\hat encia~circunscrita\end{array}}

\Large\boxed{\begin{array}{l}\sf\dfrac{a}{ sen(60^\circ)}=\dfrac{6}{ sen(45^\circ)}\\\\\sf a\cdot sen(45^\circ)=6\cdot sen(60^\circ)\\\sf a\cdot\dfrac{\sqrt{2}}{\backslash\!\!\!2}=6\cdot\dfrac{\sqrt{3}}{\backslash\!\!\!2}\\\sf a\sqrt{2}=6\sqrt{3}\\\sf a=\dfrac{6\sqrt{3}}{\sqrt{2}}\\\\\sf a=\dfrac{6\sqrt{3}}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sf a=\dfrac{\backslash\!\!\!6\sqrt{6}}{\backslash\!\!\!2}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf a=3\sqrt{6}}}}}\end{array}}

Anexos:
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