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

Answer

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free