Respuesta :

Answer:

x = 15/[tex]\sqrt{3}[/tex] and y = 30/[tex]\sqrt{3}[/tex]

Step-by-step explanation:

I just used this trick I remembered from Honors Geometry. I am now in Honors Algebra II w/ Trig. So this trick is what is used for 30-60-90 special right triangles

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