anonymous
  • anonymous
Let ABC be a triangle such that
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
a right triangle, or just some triangle?
anonymous
  • anonymous
law of cosines is my best bet
anonymous
  • anonymous
no that was a lousy bet law of sines is easier

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OpenStudier
  • OpenStudier
Law of sines: \[\frac{ \sin A }{ a }=\frac{ \sin B }{ b }\]
OpenStudier
  • OpenStudier
Do you understand?
OpenStudier
  • OpenStudier
Actually, we can compare all three: \[\frac{ \sin 30 }{ 2x+1 }=\frac{ \sin A }{ x^2+x+1 }=\frac{ \sin B }{ x^2-1 }\]
OpenStudier
  • OpenStudier
To solve for angles A and B, we know they are equal to 120 degrees (180-30).
anonymous
  • anonymous
from above we can calculate the intersection of the sides it help us to find the value of x i.e let \[^{x2+x+1= ^{x2-1}}\] because the y intersect at C i think we can solve like these
anonymous
  • anonymous
use cosine rule, cos30=(3^1/2)/2=((x^2+x+1)^2+(x^2-1)^2-(2x+1)^2)/2(x^2+x+1)(x^2-1) 3^1/2=(2x^4+2x^3-3x^2-2x+1)/(x^2+x+1)(x^2-1) by factorising the numerator 2x^4+2x^3-3x^2-2x+1=(2x^2+2x-1)(x^2-1) then solve 3^1/2(x^2+x+1)=2x^2+2x-1, its just a quadratic equation. sine rule would be too complex in this case.
anonymous
  • anonymous
\[(2x+1)^2=(x^2 + x + 1)^2+(x^2-1)^2-2(x^2+x+1)(x^2-1)\frac{\sqrt{3}}{2}\] but really i am wondering if this is a right triangle, because this looks extra annoying to solve
anonymous
  • anonymous
apparently 1 works, but that gives you nothing because then one side would be 0
anonymous
  • anonymous
\(1+\sqrt{3}\) works as well, maybe that is a good answer

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