Let $a, b$ and $c$ denote the lengths of sides of $B C, C A$ and $A B$ of $ riangle A B C$. In $ riangle A B C, \angle B A C=30^{\circ}$ and $\angle A B C=60^{\circ}$. Then $a: b: c$ is

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Let $a, b$ and $c$ denote the lengths of sides of $B C, C A$ and $A B$ of $	riangle A B C$. In $	riangle A B C, \angle B A C=30^{\circ}$ and $\angle A B C=60^{\circ}$. Then $a: b: c$ is

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Given $B C=a, C A=b$ and $A B=c$ To Find $a: b: c$

$ \begin{aligned} \because \angle A C B & =180^{\circ}-(\angle B A C+\angle A B C) \\ & =180^{\circ}-\left(30^{\circ}+60^{\circ}\right)=180^{\circ}-90^{\circ} \\ \angle A C B & =90^{\circ} \end{aligned} $ By using sine theorem, $ \begin{aligned} & \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ & \frac{\sin 30^{\circ}}{a}=\frac{\sin 60^{\circ}}{b}=\frac{\sin 90^{\circ}}{c} \\ & \Rightarrow \quad \frac{1 / 2}{a}=\frac{\sqrt{3} / 2}{b}=\frac{1}{c} \\ & \therefore \quad a: b: c=\frac{1}{2}: \frac{\sqrt{3}}{2}: 1=1: \sqrt{3}: 2 \end{aligned} $

Let $a, b$ and $c$ denote the lengths of sides of $B C, C A$ and $A B$ of $\triangle A B C$. In $\triangle A…

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