Question

Determine o valor de X nos casos :

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Answer 1

Resposta:

30* + X = 90*

x = 90* - 30*

x= 60*

b)

35* + 90* + X = 180*

125* + X = 180*

x = 180* -125*

x=55*

c)

30* + X = 90*

x = 90* - 30*

x= 60*

d)

4x - 25 + x = 90*

5x = 90* + 25

5x = 115

x=115*/5

x =23*

Explicação passo-a-passo:

espero ter ajudado.

Answer 2

$\large\boxed{\begin{array}{l}\rm a)~\sf c\acute alculo\,do\,3^o\,\hat angulo:180^\circ-(15^\circ+45^\circ)\\\sf\longrightarrow 180^\circ-60^\circ=120^\circ.\\\underline{\tt P\frak{ela\,lei\,dos\,Senos:}}\\\sf\dfrac{x}{sen(120^\circ)}=\dfrac{20}{sen(45^\circ)}\\\\\sf x\cdot sen(45^\circ)=20\cdot sen(120^\circ)\\\sf x\cdot\dfrac{\sqrt{2}}{\backslash\!\!\!2}=20\cdot\dfrac{\sqrt{3}}{\backslash\!\!\!2}\\\sf x\sqrt{2}=20\sqrt{3}\\\sf x=\dfrac{20\sqrt{3}}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\end{array}}$

$\large\boxed{\begin{array}{l}\sf x=\dfrac{20\sqrt{6}}{2}=10\sqrt{6}\end{array}}$

$\large\boxed{\begin{array}{l}\rm b)~\sf perceba\,que\,o\,3^o\,\hat angulo\,mede\,180^\circ-(30+60^\circ)\\\sf\longrightarrow 180^\circ-90^\circ=90^\circ(\acute e\,um\,tri\hat angulo\,ret\hat angulo).\\\sf portanto\\\sf tg(60^\circ)=\dfrac{x}{50}\\\\\sf x=50tg(60^\circ)\\\sf x=50\sqrt{3}\end{array}}$

$\large\boxed{\begin{array}{l}\rm c)~\sf trata-se\,do\,hex\acute agono\,inscrito.\\\sf A\,relac_{\!\!,}\tilde ao\,entre\,lado\,e\,raio\,\acute e\\\sf \sf \ell=r~~onde\,\ell~\acute e\,o\,lado\,e\,r\,o\,raio.\\\sf Na\,nossa\,figura\,x\,\acute e\,a\,medida\,do\,lado.\\\sf portanto\\\sf x=40\end{array}}$

$\large\boxed{\begin{array}{l}\rm d)~\sf trata-se\,do\,tri\hat angulo\,equil\acute atero\,inscrito.\\\sf a\,relac_{\!\!,}\tilde ao\,entre\,lado\,e\,raio\,\acute e.\\\sf \ell=r\sqrt{3}\longrightarrow r=\dfrac{\ell}{\sqrt{3}}=\dfrac{\ell\sqrt{3}}{3}.\\\sf Na\,nossa\,figura\, x\,\acute e\,o\,raio.\\\sf x=\dfrac{18\sqrt{3}}{3}\\\\\sf x=6\sqrt{3} \end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Para\,uma\,melhor\,aprofundamento}\\\underline{\rm do\,que\,foi\,dito\,aqui\,anexarei\,algumas\,figuras.}\end{array}}$