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Question

In a Δ ABC, 2a­2+4b2 + c2=4ab+2ac,then cos B is equal to


A

0

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B

1/8

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C

3/8

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D

7/8

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Solution

The correct option is D

7/8


2a2+4b2+c2=4ab+aca2+(2b)24ab+a2+c22ac=0 (a2b)2+(ac)2=0a=2b=ccosB=a2+c2b22ac=c2+c2(c2)22×c×c=2c2c242c2cosB=78


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