Respuesta :

orangesss

Hello There today we will solve your problem

Since side B is [tex]\sqrt12[/tex] we can solve that and we get

3.4641...

now we can find side c... it's a lot

[tex]c=\frac{b}{sin(b)}[/tex]

which equals 4

and

[tex]a=\sqrt c^2-b^2\\=\sqrt3.9999981349996^2-3.4641^2\\2[/tex]

[tex]\huge\red{\mid{\underline{\overline{\textbf{ANSWER}}}\mid}}[/tex]

  • Side C ā‡’ 4
  • Side A ā‡’2

x=1,

since we can use the pythagorean theorem we have x and 12

12+x= 12+x/3

1/3x+4=3

x=1

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