Let ABC be a triangle such that

Question

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 · Answer

a right triangle, or just some triangle?

anonymous · Answer

law of cosines is my best bet

anonymous · Answer

no that was a lousy bet
law of sines is easier

openstudier · Answer

Law of sines:
$$\frac{ \sin A }{ a }=\frac{ \sin B }{ b }$$

openstudier · Answer

Do you understand?

openstudier · Answer

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 · Answer

To solve for angles A and B, we know they are equal to 120 degrees (180-30).

anonymous · Answer

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 · Answer

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 · Answer

$$(2x+1)^2=(x^2 + x + 1)^2+(x^2-1)^2-2(x^2+x+1)(x^2-1)\frac{\sqrt{3}}{2}$$ but really i am wondering if this is a right triangle, because this looks extra annoying to solve

anonymous · Answer

apparently 1 works, but that gives you nothing because then one side would be 0

anonymous · Answer

$1+\sqrt{3}$ works  as well, maybe that is a good answer