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In $\triangle \mathrm{ABC}, \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in arithmetic progression and $a: c=1: 2$. If $b=4 \sqrt{3} \mathrm{~cm}$, then the area of $\triangle A B C$ (in sq. $\mathrm{cm})$ is
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The correct answer is:
$8 \sqrt{3}$
$\because A, B, C$ are in A.P. $\Rightarrow 2 B=A+C$
Now, $A+B+C=\pi \Rightarrow 3 \mathrm{~B}=\pi \Rightarrow B=\frac{\pi}{3}$
Now, $\tan \frac{(C-A)}{2}=\frac{c-a}{c+a} \cot \left(\frac{B}{2}\right)=\frac{\frac{c}{a}-1}{\frac{c}{a}+1} \cot \left(\frac{\pi}{6}\right)$ $=\frac{2-1}{2+1} \cdot \sqrt{3}=\frac{1}{3} \sqrt{3}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{C-A}{2}=\frac{\pi}{6} \Rightarrow C-A=\frac{\pi}{3}....(i)$
Also, $A+C=2 B=\frac{2 \pi}{3}...(ii)$
Solving (i) \& (ii) $\Rightarrow C=\frac{\pi}{2}$
Now, $a^2+b^2=c^2 \Rightarrow a^2+48=c^2 \quad\{\because b=4 \sqrt{3}\}$
$\Rightarrow \frac{a^2}{c^2}+\frac{48}{c^2}=1 \Rightarrow \frac{1}{4}-\frac{48}{c^2}=1 \Rightarrow c=8$
$\because \frac{a}{c}=\frac{1}{2} \Rightarrow a=4$
$c \quad 2$ Now, area of triangle $\Rightarrow \Delta=\frac{1}{2} \times a \times b=\frac{1}{2} \times 4 \times 4 \sqrt{3}=8 \sqrt{3}$
Now, $A+B+C=\pi \Rightarrow 3 \mathrm{~B}=\pi \Rightarrow B=\frac{\pi}{3}$
Now, $\tan \frac{(C-A)}{2}=\frac{c-a}{c+a} \cot \left(\frac{B}{2}\right)=\frac{\frac{c}{a}-1}{\frac{c}{a}+1} \cot \left(\frac{\pi}{6}\right)$ $=\frac{2-1}{2+1} \cdot \sqrt{3}=\frac{1}{3} \sqrt{3}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{C-A}{2}=\frac{\pi}{6} \Rightarrow C-A=\frac{\pi}{3}....(i)$
Also, $A+C=2 B=\frac{2 \pi}{3}...(ii)$
Solving (i) \& (ii) $\Rightarrow C=\frac{\pi}{2}$
Now, $a^2+b^2=c^2 \Rightarrow a^2+48=c^2 \quad\{\because b=4 \sqrt{3}\}$
$\Rightarrow \frac{a^2}{c^2}+\frac{48}{c^2}=1 \Rightarrow \frac{1}{4}-\frac{48}{c^2}=1 \Rightarrow c=8$
$\because \frac{a}{c}=\frac{1}{2} \Rightarrow a=4$
$c \quad 2$ Now, area of triangle $\Rightarrow \Delta=\frac{1}{2} \times a \times b=\frac{1}{2} \times 4 \times 4 \sqrt{3}=8 \sqrt{3}$
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