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In triangle ABC if angles A,B,C are in A.P. Then which of the following is true ?

Option: 1

b^2=a^2+c^2


Option: 2

b^2=a^2+c^2-ac


Option: 3

b^2=a^2+c^2-2ac


Option: 4

a^2=b^2+c^2-bc


Answers (1)

best_answer

A,B,C are in AP.

So, C - B = B - A

Hence, 2B=A+C

Also

\\A+B+C=180^{\circ}\\ B+2B=180^{\circ}\\ B=60^{\circ}\\ \text{Using Cosine Rule: }\cos B=\frac{a^2+c^2-b^2}{2ac}\\ \frac{1}{2}=\frac{a^2+c^2-b^2}{2ac}\\\\ a^2+c^2-b^2=ac\\ a^2+c^2-ac=b^2

 

Posted by

sudhir.kumar

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