In a triangle ABC,a=1,b=sqrt{3} and angle C=p | vedclass

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(C) In a $	riangle ABC$,it is given that $a=1$,$b=\sqrt{3}$ and $\angle C=\frac{\pi}{6}$.Using the Law of Cosines:$\cos C = \frac{a^2 + b^2 - c^2}{2a

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$1$

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(C) In a $	riangle ABC$,it is given that $a=1$,$b=\sqrt{3}$ and $\angle C=\frac{\pi}{6}$.Using the Law of Cosines:$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$\cos \frac{\pi}{6} = \frac{1^2 + (\sqrt{3})^2 - c^2}{2(1)(\sqrt{3})}$$\frac{\sqrt{3}}{2} = \frac{1 + 3 - c^2}{2\sqrt{3}}$$\frac{\sqrt{3}}{2} = \frac{4 - c^2}{2\sqrt{3}}$Multiply both sides by $2\sqrt{3}$:$3 = 4 - c^2$$c^2 = 4 - 3 = 1$Since $c$ represents the length of a side,$c = 1$.Therefore,option $C$ is correct.

In a $\triangle ABC$,$a=1$,$b=\sqrt{3}$ and $\angle C=\pi/6$. Then the measure of the third side $c=$

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