Question

5. Dado sem 75o = √2+√6
4
, determine x e y na figura

abaixo.

Answer 1

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Olhando para triângulo obtido através da altura traçada.

$\displaystyle \Delta_{\text{YHX}} \to \text{sen}(75^\circ)=\frac{\text h}{6 }\\\\\\ \Delta_{\text{YHX}} \to \frac{\sqrt{2}+\sqrt{6}}{4} = \frac{\text h}{6} \\\\\\ \Delta_{\text{YHX}} \to \text h = \frac{3(\sqrt{2}+\sqrt{6})}{2}$

Olhando para o triângulo maior :

$\displaystyle \Delta_{\text{ZHY}} \to \text{sen}(30^\circ)=\frac{\text h}{\text x} \\\\\\ \Delta_{\text{ZHY}}\to \frac{1}{2} = \frac{3(\sqrt{2}+\sqrt{6})}{2}.\frac{1}{\text x} \\\\\\ \huge\boxed{\text x = 3(\sqrt{2}+\sqrt{6}) \ \text {cm}}\checkmark$

Olhando para o triângulo amarelo, vamos aplicar lei dos senos :

$\displaystyle \frac{6}{\text{sen}(30^\circ)}=\frac{\text y}{\text{sen}(45^\circ)} \\\\\\ \text y = 6.\text{sen}(45^\circ). \frac{1}{\text{sen}(30^\circ)} \\\\\\ \text y= 6.\frac{\sqrt2}{2}.\frac{1}{\frac{1}{2}} \\\\\\ \text y = 6.\frac{\sqrt 2}{2}. 2 \\\\\\\ \huge\boxed{\text y = 6\sqrt{2} \ \text {cm}}\checkmark$