In triangle ABC, A, B and C are in arithmetic | vedclass

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(C) Since $A, B, C$ are in $A.P.$, $2B = A + C$. Given $A + B + C = \pi$, we have $3B = \pi$, so $B = \frac{\pi}{3}$. Using the Napier's analogy: $	a

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$8$

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(C) Since $A, B, C$ are in $A.P.$, $2B = A + C$. Given $A + B + C = \pi$, we have $3B = \pi$, so $B = \frac{\pi}{3}$. Using the Napier's analogy: $	an \frac{C-A}{2} = \frac{c-a}{c+a} \cot \frac{B}{2}$. Given $a:c = 1:2$, let $a = k$ and $c = 2k$. $	an \frac{C-A}{2} = \frac{2k-k}{2k+k} \cot \frac{\pi}{6} = \frac{1}{3} \cdot \sqrt{3} = \frac{1}{\sqrt{3}}$. Thus, $\frac{C-A}{2} = \frac{\pi}{6}$, which implies $C-A = \frac{\pi}{3}$. Solving $A+C = \frac{2\pi}{3}$ and $C-A = \frac{\pi}{3}$ gives $C = \frac{\pi}{2}$ and $A = \frac{\pi}{6}$. Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$. $(4\sqrt{3})^2 = a^2 + (2a)^2 - 2(a)(2a) \cos \frac{\pi}{3}$. $48 = a^2 + 4a^2 - 4a^2(\frac{1}{2}) = 3a^2$. $a^2 = 16 \Rightarrow a = 4$. Then $c = 2a = 8$. Area $= \frac{1}{2} ac \sin B = \frac{1}{2} \cdot 4 \cdot 8 \cdot \sin \frac{\pi}{3} = 16 \cdot \frac{\sqrt{3}}{2} = 8\sqrt{3} 	ext{ sq. cm}$.

In $\triangle ABC$, $A, B$ and $C$ are in arithmetic progression and $a: c = 1: 2$. If $b = 4 \sqrt{3} \text{ cm}$, then the area of $\triangle ABC$ (in $\text{sq. cm}$) is (in $\sqrt{3}$)

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