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# The angles $A,B$ and $C$ of a triangle ABC are in arithmetic progression.If $2b^2=3C^2$ determine the angle A

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

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A)
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.