If two angles of $\triangle \mathrm{ABC}$ are $\frac{\pi}{4}$ and $\frac{\pi}{3}$, then the ratio of the…
If two angles of $\triangle \mathrm{ABC}$ are $\frac{\pi}{4}$ and $\frac{\pi}{3}$, then the ratio of the smallest and greatest side is
$\sqrt{3}: \sqrt{2}$
$(\sqrt{3}-1): 1$
$(\sqrt{3}+1):(\sqrt{3}-1)$
$(\sqrt{3}+1): 1$
Solution
Two angles of triangle are $\frac{\pi}{4}$ and $\frac{\pi}{3}$ Let the third angle be $\alpha$.
$\begin{array}{l}
\therefore \frac{\pi}{4}+\frac{\pi}{3}+\alpha=\pi \\
\therefore 45^{\circ}+60^{\circ}+\alpha=180^{\circ} \Rightarrow \alpha=75^{\circ}
\end{array}$
We know that side opposite to smallest angle is the smallest side and side opposite to largest angle is the largest side.
$\therefore \frac{\mathrm{c}}{\sin 75^{\circ}}=\frac{\mathrm{a}}{\sin 45^{\circ}}$
We know that $\sin 75^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
$\therefore \frac{\mathrm{c}(2 \sqrt{2})}{\sqrt{3}+1}=\mathrm{a} \sqrt{2} \Rightarrow \frac{\mathrm{a}}{\mathrm{c}}=\frac{2}{\sqrt{3}+1}$
By rationalizing, we get $\frac{a}{c}=\frac{\sqrt{3}-1}{1}$
Note : $\sin 75^{\circ}$ can be calculated as follows :
$\begin{aligned}
\sin 75^{\circ} &=\sin \left(45^{\circ}+30^{\circ}\right) \\
&=\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)=\frac{\sqrt{3}+1}{2 \sqrt{2}}
\end{aligned}$