Question

dado o triângulo da figura, calcule X e Y.​

Answer 1

Teorema dos senos:

$\frac{x}{sen 60}=\frac{\sqrt{2} }{sen45} \\\\\frac{x}{\frac{\sqrt{3} }{2} }= \frac{\sqrt{2} }{\frac{\sqrt{2} }{2} } \\\\x\sqrt{2} =\sqrt{6}\\x=\frac{\sqrt{6} }{\sqrt{2} } \frac{\sqrt{2} }{\sqrt{2} } =\frac{\sqrt{12} }{2} \\\\x=\frac{2\sqrt{3} }2} =\sqrt{3}$

Teorema dos cossenos:

$x^{2} =(\sqrt{2} )^{2} +y^{2}-2\sqrt{2} y . cos 60\\(\sqrt{3} )^{2} =2+y^{2}-2\sqrt{2} y.\frac{1}{2} \\3=2+y^{2}-\sqrt{2} y\\y^{2}-\sqrt{2} y-1=0$

Δ=$(-\sqrt{2})^{2}-4.1.(-1)$

Δ=$2+4=6$

$\boxed{y'=\frac{\sqrt{2}+\sqrt{6} }{2}} y''=\frac{\sqrt{2}-\sqrt{6} }{2}$

é isso espero ter ajudado