If the angles of a triangle ABC are in AP,the | vedclass

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(B) Let the angles of $	riangle ABC$ be $(A-d), A, (A+d)$.Since the sum of angles in a triangle is $180^{\circ}$,we have:$(A-d) + A + (A+d) = 180^{\c

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

Vedclass · Answer

(B) Let the angles of $	riangle ABC$ be $(A-d), A, (A+d)$.Since the sum of angles in a triangle is $180^{\circ}$,we have:$(A-d) + A + (A+d) = 180^{\circ}$$3A = 180^{\circ} \Rightarrow A = 60^{\circ}$.Using the cosine rule for angle $A$:$\cos A = \frac{b^2+c^2-a^2}{2bc}$$\cos 60^{\circ} = \frac{b^2+c^2-a^2}{2bc}$$\frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$$bc = b^2+c^2-a^2$$a^2 = b^2+c^2-bc$.Thus,the correct relation is $a^2 = b^2+c^2-bc$.

If the angles of a $\triangle ABC$ are in $AP$,then

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