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Question

If the angles A<B<C of a ABC are in A. P. then


A

c2=a2+b2ab2

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B

c2=a2+b2

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C

a2=b2+c2bc

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D

b2=a2+c2ac2

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Solution

The correct option is D

b2=a2+c2ac2


Explanation for Correct Option:

Step 1: The Sum of all angles of the triangle is 180°.

So, A+B+C=180°.....(i)

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

So, 2B=A+C.....(ii)

Step 2: Putting the value of equation (ii) in equation (i) We get,

3B=180°orB=60°

Now,

b2=a2+c22accosBb2=a2+c22accos60°b2=a2+c2ac

Hence option (3) is Correct


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