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anonymous
 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?
anonymous
 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a right triangle, or just some triangle?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0law of cosines is my best bet

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no that was a lousy bet law of sines is easier

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Law of sines: \[\frac{ \sin A }{ a }=\frac{ \sin B }{ b }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, we can compare all three: \[\frac{ \sin 30 }{ 2x+1 }=\frac{ \sin A }{ x^2+x+1 }=\frac{ \sin B }{ x^21 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To solve for angles A and B, we know they are equal to 120 degrees (18030).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from above we can calculate the intersection of the sides it help us to find the value of x i.e let \[^{x2+x+1= ^{x21}}\] because the y intersect at C i think we can solve like these

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use cosine rule, cos30=(3^1/2)/2=((x^2+x+1)^2+(x^21)^2(2x+1)^2)/2(x^2+x+1)(x^21) 3^1/2=(2x^4+2x^33x^22x+1)/(x^2+x+1)(x^21) by factorising the numerator 2x^4+2x^33x^22x+1=(2x^2+2x1)(x^21) then solve 3^1/2(x^2+x+1)=2x^2+2x1, its just a quadratic equation. sine rule would be too complex in this case.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(2x+1)^2=(x^2 + x + 1)^2+(x^21)^22(x^2+x+1)(x^21)\frac{\sqrt{3}}{2}\] but really i am wondering if this is a right triangle, because this looks extra annoying to solve

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0apparently 1 works, but that gives you nothing because then one side would be 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(1+\sqrt{3}\) works as well, maybe that is a good answer
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