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 $a=x^2+x+1,b=x^2-1$and $c=2x+1$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

$1+\sqrt{3}$

Explanation for the correct option:

Finding the value of $x$:

The given angle,

$\angle ACB=\frac{\pi }{6}$

And the sides,

$a=x^2+x+1,b=x^2-1$ and $c=2x+1$

We know that according to cosine rule,

$\cos \angle ACB=\frac{(a^2+b^2–c^2)}{2ab}$

$$\begin{gathered} \therefore \cos \frac{\mathrm{\pi }}{6}=\frac{\left[{(x^2+x+1)}^2+{(x^2–1)}^2–{(2x+1)}^2\right]}{\left[2(x^2+x+1)(x^2–1)\right]} \\ \Rightarrow \sqrt{3}=\frac{(x^4+{\left(2x\right)}^2+1^2+2x^3+2x+2x^2)+\left(x^4+1-2x^2\right)-\left({\left(2x\right)}^2+1+4x\right)}{(x^4-x^2+x^3-x+x^2-1)} \\ \Rightarrow \sqrt{3}=\frac{(2x^2+2x–1)}{(x^2+x+1)} \\ \mathrm{Now}, \\ \Rightarrow \sqrt{3}(x^2+x+1)=(2x^2+2x–1) \\ \mathrm{Solving}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{we}\mathrm{get} \\ \Rightarrow (\sqrt{3}–2)x^2+(\sqrt{3}–2)x+(\sqrt{3}+1)=0 \end{gathered}$$

Solving the above equation by using quadratic formula,

$$\begin{gathered} \Rightarrow x=\frac{\left[(2–\sqrt{3})\pm \sqrt{3}\right]}{2(\surd 3–2)} \\ \Rightarrow x=\frac{\left[(2–\sqrt{3})+\sqrt{3}\right]}{2(\surd 3–2)}or\frac{\left[(2–\sqrt{3})-\sqrt{3}\right]}{2(\surd 3–2)} \\ \Rightarrow x=\frac{1}{(\surd 3–2)}or\frac{(1–\sqrt{3})}{(\surd 3–2)} \\ \Rightarrow x=\frac{1}{(\surd 3–2)}\left[\frac{(\surd 3+2)}{(\surd 3+2)}\right]or\frac{(1–\sqrt{3})}{(\surd 3–2)}\left[\frac{(\surd 3+2)}{(\surd 3+2)}\right] \\ \Rightarrow x=-(2+\surd 3)orx=1+\surd 3 \end{gathered}$$

Therefore,

$x=1+\sqrt{3}$ as $x>0$
Hence, the correct option is (B).