In a triangle ABC,if angle A = 60^{circ},then | vedclass

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(A) Given $\angle A = 60^{\circ}$. By the Cosine Rule,$a^2 = b^2 + c^2 - 2bc \cos A$. Since $\cos 60^{\circ} = \frac{1}{2}$,we have $a^2 = b^2 + c^2 -

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$3bc$

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(A) Given $\angle A = 60^{\circ}$. By the Cosine Rule,$a^2 = b^2 + c^2 - 2bc \cos A$. Since $\cos 60^{\circ} = \frac{1}{2}$,we have $a^2 = b^2 + c^2 - 2bc \left(\frac{1}{2}\right) = b^2 + c^2 - bc$. Now,consider the expression $(a+b+c)(b+c-a)$. This can be rewritten as $((b+c)+a)((b+c)-a) = (b+c)^2 - a^2$. Expanding this,we get $b^2 + c^2 + 2bc - a^2$. Substituting $a^2 = b^2 + c^2 - bc$,we get: $b^2 + c^2 + 2bc - (b^2 + c^2 - bc) = b^2 + c^2 + 2bc - b^2 - c^2 + bc = 3bc$.

In a $\triangle ABC$,if $\angle A = 60^{\circ}$,then $(a+b+c)(b+c-a) =$

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