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If two angles of $\triangle \mathrm{ABC}$ are $\frac{\pi}{4}$ and $\frac{\pi}{3}$, then the ratio of the smallest and greatest sides are
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Verified Answer
The correct answer is:
$(\sqrt{3}-1): 1$
Let three angles of the triangle be given as $\mathrm{A}=\frac{\pi}{4}, \mathrm{~B}=\frac{\pi}{3}$ and $\mathrm{C}=\frac{\pi}{4}+\frac{\pi}{6}$
Let a, b, c be the opposite to angles A, B, C respectively.
As $\frac{\sin A}{a}=\frac{\sin C}{c}$, we get
$$
\text { Required Ratio }=\frac{\mathrm{a}}{\mathrm{c}}=\frac{\sin \mathrm{A}}{\sin \mathrm{C}}
$$
$$
=\frac{\sin \left(\frac{\pi}{4}\right)}{\sin \left(\frac{\pi}{4}+\frac{\pi}{6}\right)}
$$
$\begin{aligned} & =\frac{\sin \frac{\pi}{4}}{\sin \frac{\pi}{4} \cos \frac{\pi}{6}+\cos \frac{\pi}{4} \sin \frac{\pi}{6}} \\ & =\frac{2}{\sqrt{3}+1} \\ & =\frac{2(\sqrt{3}-1)}{2} \\ & =(\sqrt{3}-1): 1\end{aligned}$
Let a, b, c be the opposite to angles A, B, C respectively.
As $\frac{\sin A}{a}=\frac{\sin C}{c}$, we get
$$
\text { Required Ratio }=\frac{\mathrm{a}}{\mathrm{c}}=\frac{\sin \mathrm{A}}{\sin \mathrm{C}}
$$
$$
=\frac{\sin \left(\frac{\pi}{4}\right)}{\sin \left(\frac{\pi}{4}+\frac{\pi}{6}\right)}
$$
$\begin{aligned} & =\frac{\sin \frac{\pi}{4}}{\sin \frac{\pi}{4} \cos \frac{\pi}{6}+\cos \frac{\pi}{4} \sin \frac{\pi}{6}} \\ & =\frac{2}{\sqrt{3}+1} \\ & =\frac{2(\sqrt{3}-1)}{2} \\ & =(\sqrt{3}-1): 1\end{aligned}$
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