Question
In a $\triangle\text{ABC},$ if $\text{b}=\sqrt{3},$ c = 1 and $\angle\text{A}=30^{\circ},$ find a.

Answer

We know, $\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$ $\Rightarrow\cos30=\frac{\sqrt{3^2}+1^2-\text{a}^2}{2\times\sqrt{3}\times1}$ $\Rightarrow\frac{\sqrt{3}}{2}=\frac{3+1-\text{a}^2}{2\sqrt{3}}$ $\Rightarrow2\sqrt{3}\times\sqrt{3}=2(4-\text{a}^2)$ $\Rightarrow3=4-\text{a}^2$ $\Rightarrow\text{a}^2=1$ $\Rightarrow\text{a}=1$

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