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  • 3 years ago

Let ABC be a triangle such that <ACB = 30 let a,b,c denote the lenght of the sides oppositr to A,B,C. The Values for which a=x^2 + x + 1 , b = x^2 -1 and c= 2x +1 is?

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  1. anonymous
    • 3 years ago
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    a right triangle, or just some triangle?

  2. anonymous
    • 3 years ago
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    law of cosines is my best bet

  3. anonymous
    • 3 years ago
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    no that was a lousy bet law of sines is easier

  4. OpenStudier
    • 3 years ago
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    Law of sines: \[\frac{ \sin A }{ a }=\frac{ \sin B }{ b }\]

  5. OpenStudier
    • 3 years ago
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    Do you understand?

  6. OpenStudier
    • 3 years ago
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    Actually, we can compare all three: \[\frac{ \sin 30 }{ 2x+1 }=\frac{ \sin A }{ x^2+x+1 }=\frac{ \sin B }{ x^2-1 }\]

  7. OpenStudier
    • 3 years ago
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    To solve for angles A and B, we know they are equal to 120 degrees (180-30).

  8. neba.zi
    • 3 years ago
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    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

  9. melanka
    • 3 years ago
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    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.

  10. anonymous
    • 3 years ago
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    \[(2x+1)^2=(x^2 + x + 1)^2+(x^2-1)^2-2(x^2+x+1)(x^2-1)\frac{\sqrt{3}}{2}\] but really i am wondering if this is a right triangle, because this looks extra annoying to solve

  11. anonymous
    • 3 years ago
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    apparently 1 works, but that gives you nothing because then one side would be 0

  12. anonymous
    • 3 years ago
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    \(1+\sqrt{3}\) works as well, maybe that is a good answer

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