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Two angles of a triangle are $\frac{\pi}{6}$ and $\frac{\pi}{4}$ and the length of the included side is $(\sqrt{3}+1) \mathrm{cm}$. The area of the triangle is
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The correct answer is:
$\frac{\sqrt{3}+1}{2} \mathrm{~cm}^2$
$A=\pi-\frac{\pi}{4}-\frac{\pi}{6}=180^{\circ}-45^{\circ}-30^{\circ}=105^{\circ}$
$\therefore \frac{\sqrt{3}+1}{\sin 105^{\circ}}=\frac{c}{\sin \frac{\pi}{6}}$ i.e. $\frac{C}{\sin 30^{\circ}}$
$\therefore C=\frac{1}{2} \cdot \frac{\sqrt{3}+1}{\sin \left(60^{\circ}+45^{\circ}\right)}$
$=\frac{\sqrt{3}+1}{2\left(\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right)}$
$=\frac{(\sqrt{3}+1)}{2 \cdot(\sqrt{3}+1)} \cdot 2 \sqrt{2}=\sqrt{2}$
$\therefore$ reqd. area $=\frac{1}{2}$ ca $\sin B$
$=\frac{1}{2} \sqrt{2}(\sqrt{3}+1) \sin \frac{\pi}{4}$
$=\frac{\sqrt{2}}{2}(\sqrt{3}+1) \cdot \frac{1}{\sqrt{2}}$
$=\frac{\sqrt{3}+1}{2} \mathrm{~cm}^2$
$\therefore \frac{\sqrt{3}+1}{\sin 105^{\circ}}=\frac{c}{\sin \frac{\pi}{6}}$ i.e. $\frac{C}{\sin 30^{\circ}}$
$\therefore C=\frac{1}{2} \cdot \frac{\sqrt{3}+1}{\sin \left(60^{\circ}+45^{\circ}\right)}$
$=\frac{\sqrt{3}+1}{2\left(\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right)}$
$=\frac{(\sqrt{3}+1)}{2 \cdot(\sqrt{3}+1)} \cdot 2 \sqrt{2}=\sqrt{2}$
$\therefore$ reqd. area $=\frac{1}{2}$ ca $\sin B$
$=\frac{1}{2} \sqrt{2}(\sqrt{3}+1) \sin \frac{\pi}{4}$
$=\frac{\sqrt{2}}{2}(\sqrt{3}+1) \cdot \frac{1}{\sqrt{2}}$
$=\frac{\sqrt{3}+1}{2} \mathrm{~cm}^2$
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