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Let $a, b$ and $c$ denote the lengths of sides of $B C, C A$ and $A B$ of $\triangle A B C$. In $\triangle A B C, \angle B A C=30^{\circ}$ and $\angle A B C=60^{\circ}$.
Then $a: b: c$ is
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Then $a: b: c$ is
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Verified Answer
The correct answer is:
$1: \sqrt{3}: 2$
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,
$$
\begin{aligned}
& \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\
& \frac{\sin 30^{\circ}}{a}=\frac{\sin 60^{\circ}}{b}=\frac{\sin 90^{\circ}}{c} \\
& \Rightarrow \quad \frac{1 / 2}{a}=\frac{\sqrt{3} / 2}{b}=\frac{1}{c} \\
& \therefore \quad a: b: c=\frac{1}{2}: \frac{\sqrt{3}}{2}: 1=1: \sqrt{3}: 2
\end{aligned}
$$
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,
$$
\begin{aligned}
& \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\
& \frac{\sin 30^{\circ}}{a}=\frac{\sin 60^{\circ}}{b}=\frac{\sin 90^{\circ}}{c} \\
& \Rightarrow \quad \frac{1 / 2}{a}=\frac{\sqrt{3} / 2}{b}=\frac{1}{c} \\
& \therefore \quad a: b: c=\frac{1}{2}: \frac{\sqrt{3}}{2}: 1=1: \sqrt{3}: 2
\end{aligned}
$$
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