In a triangle ABC, the angles A, B, C are in A.P. Show that 2 cos [(A-C)/2] = a+c/sqrt(a^2-ac+c^2)

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Arun Kumar · Answer

Hi Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

Akshay Chaudhari · Answer

By cosine law i.e. 2accosB=a^2+c^2-b^2 RHS=a+c/b By sine law i.e. a=2RSinA,and similarly Further applying Formula of SinA + SinC in LHS Hence LHS=RHS

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In a triangle ABC, the angles A, B, C are in A.P. Show that 2 cos [(A-C)/2] = a+c/sqrt(a^2-ac+c^2)

In a triangle ABC, the angles A, B, C are in A.P. Show that 2 cos [(A-C)/2] = a+c/sqrt(a^2-ac+c^2)

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In a triangle ABC, the angles A, B, C are in A.P. Show that 2 cos [(A-C)/2] = a+c/sqrt(a^2-ac+c^2)

In a triangle ABC, the angles A, B, C are in A.P. Show that 2 cos [(A-C)/2] = a+c/sqrt(a^2-ac+c^2)

2 Answers

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