Question

Calcule o raio da circunferência circunscrita ao triângulo nos casos:

Lembrando darei melhor resposta ✓​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{sen\:{\^A} = \dfrac{a}{2R}}$

$\mathsf{sen\:45\textdegree = \dfrac{16}{2R}}$

$\mathsf{\dfrac{\sqrt{2}}{2} = \dfrac{16}{2R}}$

$\mathsf{R = \dfrac{16}{\sqrt{2}}}$

$\boxed{\boxed{\mathsf{R = 8\sqrt{2}}}}$

$\mathsf{sen\:{\^A} = \dfrac{a}{2R}}$

$\mathsf{sen\:30\textdegree = \dfrac{15}{2R}}$

$\mathsf{\dfrac{1}{2} = \dfrac{15}{2R}}$

$\boxed{\boxed{\mathsf{R = 15}}}$

$\mathsf{sen\:{\^A} = \dfrac{a}{2R}}$

$\mathsf{sen\:120\textdegree = \dfrac{24}{2R}}$

$\mathsf{\dfrac{\sqrt{3}}{2} = \dfrac{24}{2R}}$

$\mathsf{R = \dfrac{24}{\sqrt{3}}}$

$\boxed{\boxed{\mathsf{R = 8\sqrt{3}}}}$

Answer 2

$\large\boxed{\begin{array}{l}\rm a)~\sf trata-se\,do\,quadrado\,inscrito.\\\sf a \,relac_{\!\!,}\tilde ao\,entre\,lado\,e\,raio\,\acute e\\\sf \ell =r\sqrt{2}\\\sf r=\dfrac{\ell}{\sqrt{2}}=\dfrac{\ell\sqrt{2}}{2}\\\\\sf r=\dfrac{16\sqrt{2}}{2}=8\sqrt{2} \end{array}}$

$\large\boxed{\begin{array}{l}\rm b)~\sf trata-se\,do\,hex\acute agono\,inscrito.\\\sf A\,relac_{\!\!,}\tilde ao\,entre\,lado\,e\,raio\,\acute e:\\\sf \ell=r\\\sf portanto\\\sf r=15\end{array}}$

$\large\boxed{\begin{array}{l}\rm c)~\sf trata-se\,do\,tri\hat angulo\,equil\acute atero\,inscrito.\\\sf A\,relac_{\!\!,}\tilde ao\,entre\,lado\,e\,raio\,\acute e:\\\sf \ell=r\sqrt{3}\implies r=\dfrac{\ell}{\sqrt{3}}=\dfrac{\ell\sqrt{3}}{3}\\\\\sf r=\dfrac{24\sqrt{3}}{3}\\\\\sf r=8\sqrt{3}\end{array}}$