Question

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Answer 1

Sejam os ângulos internos do triângulo $ABC$:

$\begin{array}{cl} \bullet&\mathrm{med(A\widehat{B}C)}=30^{\circ}\\ \\ \bullet&\mathrm{med(B\widehat{C}A)}=\alpha\\ \\ \bullet&\mathrm{med(C\widehat{A}B)}=150^{\circ}-\alpha \end{array}$


E a medida do lado desconhecido $\overline{BC}$:

$\mathrm{med}(\overline{BC})=x$


1) Pela Lei dos Senos no triângulo $ABC$, temos

$\dfrac{2\sqrt{7}}{\mathrm{sen\,}30^{\circ}}=\dfrac{2\sqrt{3}}{\mathrm{sen\,}\alpha}\\ \\ \dfrac{2\sqrt{7}}{^{1}\!\!\!\diagup\!\!_{2}}=\dfrac{2\sqrt{3}}{\mathrm{sen\,}\alpha}\\ \\ 2\sqrt{7}\cdot \mathrm{sen\,}\alpha=\dfrac{1}{\diagup\!\!\!\! 2}\cdot \diagup\!\!\!\! 2\sqrt{3}\\ \\ 2\sqrt{7}\cdot \mathrm{sen\,}\alpha=\sqrt{3}\\ \\ \boxed{\mathrm{sen\,}\alpha=\dfrac{\sqrt{3}}{2\sqrt{7}}}$


Podemos racionalizar o denominador, multiplicando o numerador e o denominador por $\sqrt{7}$:

$\mathrm{sen\,}\alpha=\dfrac{\sqrt{3}\cdot \sqrt{7}}{2\sqrt{7}\cdot \sqrt{7}}\\ \\ \mathrm{sen\,}\alpha=\dfrac{\sqrt{3\cdot 7}}{2\cdot \left(\sqrt{7} \right )^{2}}\\ \\ \mathrm{sen\,}\alpha=\dfrac{\sqrt{21}}{2\cdot 7}\\ \\ \boxed{\mathrm{sen\,}\alpha=\dfrac{\sqrt{21}}{14}}$


Vamos encontrar o cosseno de $\alpha$, utilizando a Relação Trigonométrica Fundamental:

$\mathrm{sen^{2\,}}\alpha+\cos^{2}\alpha=1\\ \\ \cos^{2}\alpha=1-\mathrm{sen^{2\,}}\alpha\\ \\ \cos^{2}\alpha=1-\left(\dfrac{\sqrt{3}}{2\sqrt{7}} \right )^{2}\\ \\ \cos^{2}\alpha=1-\dfrac{\left(\sqrt{3} \right )^{2}}{\left(2\sqrt{7} \right )^{2}}\\ \\ \cos^{2}\alpha=1-\dfrac{3}{2^{2}\cdot \left(\sqrt{7} \right )^{2}}\\ \\ \cos^{2}\alpha=1-\dfrac{3}{4\cdot 7}\\ \\ \cos^{2}\alpha=1-\dfrac{3}{28}\\ \\ \cos^{2}\alpha=\dfrac{28-3}{28}\\ \\ \cos^{2}\alpha=\dfrac{25}{28}\\ \\ \cos \alpha=\sqrt{\dfrac{25}{28}}\\ \\ \cos \alpha=\dfrac{\sqrt{25}}{\sqrt{28}}\\ \\ \cos \alpha=\dfrac{5}{\sqrt{2^{2}\cdot 7}}\\ \\ \cos \alpha=\dfrac{5}{\sqrt{2^{2}}\cdot\sqrt{7}}\\ \\ \boxed{\cos \alpha=\dfrac{5}{2\sqrt{7}}}$


Podemos racionalizar o denominador, multiplicando o numerador e o denominador por $\sqrt{7}$:

$\cos \alpha=\dfrac{5\cdot \sqrt{7}}{2\sqrt{7}\cdot \sqrt{7}}\\ \\ \cos \alpha=\dfrac{5\sqrt{7}}{2\cdot \left(\sqrt{7} \right )^{2}}\\ \\ \cos \alpha=\dfrac{5\sqrt{7}}{2\cdot 7}\\ \\ \boxed{\cos \alpha=\dfrac{5\sqrt{7}}{14}}$


2) Utilizaremos a Lei dos Cossenos para o lado $\overline{BC}$. Para isso, devemos encontrar a medida do cosseno do ângulo oposto a este lado:

$\cos (150^{\circ}-\alpha)=\cos 150^{\circ}\cdot \cos \alpha+\mathrm{sen\,}150^{\circ}\cdot \mathrm{sen\,}\alpha\\ \\ \cos (150^{\circ}-\alpha)=-\dfrac{\sqrt{3}}{2}\cdot \dfrac{5\sqrt{7}}{14}+\dfrac{1}{2}\cdot \dfrac{\sqrt{21}}{14}\\ \\ \cos (150^{\circ}-\alpha)=\dfrac{-5\sqrt{21}}{28}+\dfrac{\sqrt{21}}{28}\\ \\ \cos (150^{\circ}-\alpha)=\dfrac{-5\sqrt{21}+\sqrt{21}}{28}\\ \\ \cos (150^{\circ}-\alpha)=\dfrac{-4\sqrt{21}}{28}\\ \\ \cos (150^{\circ}-\alpha)=\dfrac{-\diagup\!\!\!\! 4\sqrt{21}}{\diagup\!\!\!\! 4\cdot 7}\\ \\ \boxed{\cos (150^{\circ}-\alpha)=-\dfrac{\sqrt{21}}{7}}$


Aplicando a Lei dos Cossenos ao lado $\overline{BC}$, temos:

$x^{2}=\left(2\sqrt{3} \right )^{2}+\left(2\sqrt{7} \right )^{2}-2\cdot 2\sqrt{3}\cdot 2\sqrt{7}\cdot \cos (150^{\circ}-\alpha)\\ \\ x^{2}=2^{2}\cdot \left(\sqrt{3} \right )^{2}+2^{2}\cdot \left(\sqrt{7} \right )^{2}-8\sqrt{21}\cdot \left(-\dfrac{\sqrt{21}}{7} \right )\\ \\ x^{2}=4\cdot 3+4\cdot 7+\dfrac{8\sqrt{21}\cdot \sqrt{21}}{7}\\ \\ x^{2}=12+28+\dfrac{8\cdot \left(\sqrt{21} \right )^{2}}{7}\\ \\ x^{2}=40+\dfrac{8\cdot 21}{7}\\ \\ x^{2}=40+\dfrac{8\cdot 3\cdot \diagup\!\!\!\! 7}{\diagup\!\!\!\! 7}\\ \\ x^{2}=40+8\cdot 3\\ \\ x^{2}=40+24\\ \\ x^{2}=64\\ \\ x=\sqrt{64}\\ \\ x=8\text{ cm}$