Question

Let ABC be a triangle such that $\angle ACB=\frac{\pi }{6}$ and let $a,b,$ and $c$ denote the lengths of the sides opposite to $A,B,$ and $C$, respectively. The value(s) of $x,$ for which $a=x^2+1,b=x^2-1$ and $c=2x+1$ are valid, is (are)

• A. $-(2+\sqrt{3})$
• B. $1+\sqrt{3}$
• C. $2+\sqrt{3}$
• D. $4\sqrt{3}$

Answer 1

The correct option is B $1+\sqrt{3}$

$\frac{\sqrt{3}}{2}=\frac{(x^2+x+1{)}^2+(x^2-1{)}^2-(2x+1{)}^2}{2(x^2+x+1)(x^2-1)}$

$(x+2)(x+1)(x-1)x+{(x^2-1)}^2=\sqrt{3}(x^2+x+1){(x}^2-1)$

$(2-\sqrt{3})x^2+(2-\sqrt{3})x-(\sqrt{3}+1)=0$

$x=(1+\sqrt{3})$