In triangle ABC, with usual notations, if the | vedclass

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(A) Given that angles $A, B, C$ are in Arithmetic Progression, let $A = a-d, B = a, C = a+d$.Since $A+B+C = \pi$, we have $3a = \pi$, so $a = \frac{\p

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$\frac{5\pi}{12}$

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(A) Given that angles $A, B, C$ are in Arithmetic Progression, let $A = a-d, B = a, C = a+d$.Since $A+B+C = \pi$, we have $3a = \pi$, so $a = \frac{\pi}{3}$. Thus, $B = 60^\circ$.Using the Sine Rule, $\frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, so $b = 2R \sin B$ and $c = 2R \sin C$.The given condition is $3c^2 = 2b^2$, which implies $3(2R \sin C)^2 = 2(2R \sin B)^2$.$3 \sin^2 C = 2 \sin^2 60^\circ = 2 	imes (\frac{\sqrt{3}}{2})^2 = 2 	imes \frac{3}{4} = \frac{3}{2}$.$\sin^2 C = \frac{1}{2}$, so $\sin C = \frac{1}{\sqrt{2}}$. Thus, $C = 45^\circ$ or $C = 135^\circ$.If $C = 45^\circ$, then $A = 180^\circ - 60^\circ - 45^\circ = 75^\circ = \frac{5\pi}{12}$.If $C = 135^\circ$, then $A = 180^\circ - 60^\circ - 135^\circ = -15^\circ$, which is impossible.Therefore, $A = \frac{5\pi}{12}$.

In $\triangle ABC$, with usual notations, if the angles are in Arithmetic Progression and $3c^2 = 2b^2$, then the measure of angle $A$ is

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