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Q.

Let sin Asin B=sinA-CsinC-B, where A,B,C are angles of a triangle ABC. If the lengths of the sides opposite to these angles are  a, b, c respectively, then

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a

b2, c2, a2 are in A.P.

b

b2-a2=a2+c2.

c

c2, a2, b2 are in A.P.

d

a2, b2, c2 are in A.P

answer is A.

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Detailed Solution

We have sin Asin B=sinA-CsinC-B sin Asin B=sin A. cos C-cos A, sin CsinC.cosB-cosC.sin Bab= aa2+b2-c2ab-b2+c2-a22bc.cca2+c2-b22ac-a2+b2-c22ab.b Using sine and cosine rule ab=aba2+b2-c2-b2-c2+a2a2+c2-b2-a2-b2+c2 c2-b2=a2-c22c2=b2+a2b2, c2, a2 are in A.P. 

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