In a triangle ABC,if b=2, c=3 and angle B=fra | vedclass

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(C) Given,$b = 2$,$c = 3$,and $\angle B = \frac{\pi}{6}$.Using the cosine rule in $	riangle ABC$:$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$Substituting t

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$a^2-3 \sqrt{3} a+5=0$

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(C) Given,$b = 2$,$c = 3$,and $\angle B = \frac{\pi}{6}$.Using the cosine rule in $	riangle ABC$:$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$Substituting the given values:$\cos \frac{\pi}{6} = \frac{a^2 + 3^2 - 2^2}{2 \cdot a \cdot 3}$$\frac{\sqrt{3}}{2} = \frac{a^2 + 9 - 4}{6a}$$\frac{\sqrt{3}}{2} = \frac{a^2 + 5}{6a}$Multiplying both sides by $6a$:$3\sqrt{3} a = a^2 + 5$Rearranging the terms,we get:$a^2 - 3\sqrt{3} a + 5 = 0$

In a $\triangle ABC$,if $b=2, c=3$ and $\angle B=\frac{\pi}{6}$,then $a$ satisfies the equation

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