Question

NO TRIÂNGULO SEGUINTE CALCULAR AS MEDIDAS DE X:​

Answer 1

• Lei dos senos

$\frac{sen135}{x} = \frac{sen30}{ \sqrt{2} }$

$\frac{ \frac{ \sqrt{2} }{2} }{x} = \frac{ \frac{1}{2} }{ \sqrt{2} }$

$\frac{1}{2} x = \frac{ \sqrt{2} }{2} \times \sqrt{2}$

$\frac{1}{2} x = \frac{ \sqrt{4} }{2}$

$x = \frac{ \frac{ \sqrt{4} }{2} }{ \frac{1}{2} }$

$x = \frac{ \sqrt{4} }{2} \times \frac{2}{1}$

$x = \frac{2 \sqrt{4} }{2}$

$x = \sqrt{4}$

$\boxed{x=2}$

Espero ter ajudado!

Answer 2

$\large\boxed{\begin{array}{l}\sf \alpha=180^\circ-(15^\circ+30^\circ)\\\sf \alpha=180^\circ-45^\circ\\\sf \alpha=135^\circ\\\underline{\rm Pela\,lei\,dos\,senos\,temos:}\\\sf \dfrac{x}{sen(135^\circ)}=\dfrac{ \sqrt{2}}{sen(30^\circ)}\\\\\sf x\cdot sen(30^\circ)=\sqrt{2}\cdot sen(135^\circ)\\\sf x\cdot\dfrac{1}{\backslash\!\!\!2}=\sqrt{2}\cdot\dfrac{\sqrt{2}}{\backslash\!\!\!2}\\\\\sf x=\sqrt{2}\cdot\sqrt{2}\\\sf x=\sqrt{4}\\\sf x=2\end{array}}$