$(a)\;90^{\large\circ}\qquad(b)\;45^{\large\circ}\qquad(c)\;75^{\large\circ}\qquad(d)\;60^{\large\circ}$

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Step 1:

Since $A,B,C$ are in A.P

$2B=A+C$

Also,$A+B+C=180^{\large\circ}$

$2B+B=18^{\large\circ}$

$3B=180^{\large\circ}$

$B=\large\frac{180^{\large\circ}}{3}$

$B=60^{\large\circ}$

Step 2:

Given :

$2b^2=3C^2$

$2k^2\sin^2B=3k^2\sin^2C$

$2\sin^260^{\large\circ}=3\sin^2C$

$2.\large\frac{3}{4}$$=3\sin^2C$

$\sin^2C=\large\frac{1}{2}$

$\sin C=\large\frac{1}{\sqrt 2}$

$C=\sin^{-1}\large\frac{1}{\sqrt 2}$

$C=45^{\large\circ}$

Step 3:

Hence $A=180^{\large\circ}-(B+C)$

On substituting B & C we get,

$A=180^{\large\circ}-(60^{\large\circ}+45^{\large\circ})$

$\;\;\;=75^{\large\circ}$

Hence (c) is the correct answer.

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