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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?
a right triangle, or just some triangle?
law of cosines is my best bet
no that was a lousy bet law of sines is easier
Law of sines: \[\frac{ \sin A }{ a }=\frac{ \sin B }{ b }\]
Actually, we can compare all three: \[\frac{ \sin 30 }{ 2x+1 }=\frac{ \sin A }{ x^2+x+1 }=\frac{ \sin B }{ x^2-1 }\]
To solve for angles A and B, we know they are equal to 120 degrees (180-30).
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
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.
\[(2x+1)^2=(x^2 + x + 1)^2+(x^2-1)^2-2(x^2+x+1)(x^2-1)\frac{\sqrt{3}}{2}\] but really i am wondering if this is a right triangle, because this looks extra annoying to solve
apparently 1 works, but that gives you nothing because then one side would be 0
\(1+\sqrt{3}\) works as well, maybe that is a good answer