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If the angles of a triangle are $30^{\circ}$ and $45^{\circ}$ and the included
side is $(\sqrt{3}+1)$, then what is the area of the tringle?
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side is $(\sqrt{3}+1)$, then what is the area of the tringle?
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}+1}{2}$
From $\Delta \mathrm{ADB}, \mathrm{AD}=\mathrm{BD}=\mathrm{x}$
In $\Delta \mathrm{ADC}$
$\tan 30^{\circ}=\frac{\mathrm{x}}{\sqrt{3}+1-\mathrm{x}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{x}}{\sqrt{3}+1-\mathrm{x}} \Rightarrow \sqrt{3} \mathrm{x}=\sqrt{3}+1-\mathrm{x}$
$\Rightarrow(\sqrt{3}+1) \mathrm{x}=\sqrt{3}+1$
$x=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1$
Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times(\sqrt{3}+1) \times 1=\frac{\sqrt{3}+1}{2}$
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