In a $\triangle A B C$, if $b=2, c=3$ and $\angle B=\frac{\pi}{6}$, then $a$ satisfies the equation
- $a^2+3 \sqrt{3 a}+5=0$
- $a^2+3 \sqrt{3 a}-5=0$
- $a^2-3 \sqrt{3 a}+5=0$
- $\sqrt{3} a^2+3 a+5=0$
Solution

By using cosine theorem in $\triangle A B C$, $ \begin{aligned} & \cos B=\frac{a^2+c^2-b^2}{2 a c} \\ & \cos \frac{\pi}{6}=\frac{a^2+3^2-2^2}{2 \cdot a \cdot 3} \Rightarrow \frac{\sqrt{3}}{2}=\frac{a^2+5}{6 a} \\ \Rightarrow & a^2-3 \sqrt{3} a+5=0 \end{aligned} $
Asked in: AP EAMCET 2021 (25 Aug Shift 2)