In a \(\triangle A B C, a=1, b=\sqrt{3}\) and \(\angle C=\pi / 6\). Then the measure of the third side \(c=\)
In a \(\triangle A B C, a=1, b=\sqrt{3}\) and \(\angle C=\pi / 6\). Then the measure of the third side \(c=\)
4
3
1
2
Solution
In a \(\triangle A B C\), it is given that, \(a=1, b=\sqrt{3}\) and \(\angle C=\frac{\pi}{6}\), then by cosine law, we have
\(\begin{aligned}
\cos \frac{\pi}{6} & =\frac{a^2+b^2-c^2}{2 a b} \\
\Rightarrow \quad \frac{\sqrt{3}}{2} & =\frac{1+3-c^2}{2 \sqrt{3}} \Rightarrow 3=4-c^2 \\
\Rightarrow c^2=1 \Rightarrow c & = \pm 1 \Rightarrow c=1 \quad \{\because c=\text { length of side}\}
\end{aligned}\)
Hence, option (c) is correct.