In triangle ABC,if a=13, b=8, c=7,then cos(B+ | vedclass

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(D) In any triangle $ABC$,the sum of angles is $A+B+C = 180^{\circ}$.Therefore,$B+C = 180^{\circ} - A$.Thus,$\cos(B+C) = \cos(180^{\circ} - A) = -\cos

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$\frac{1}{2}$

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(D) In any triangle $ABC$,the sum of angles is $A+B+C = 180^{\circ}$.Therefore,$B+C = 180^{\circ} - A$.Thus,$\cos(B+C) = \cos(180^{\circ} - A) = -\cos A$.Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting the given values $a=13, b=8, c=7$:$\cos A = \frac{8^2 + 7^2 - 13^2}{2 	imes 8 	imes 7} = \frac{64 + 49 - 169}{112} = \frac{113 - 169}{112} = \frac{-56}{112} = -\frac{1}{2}$.Finally,$\cos(B+C) = -\cos A = -(-\frac{1}{2}) = \frac{1}{2}$.

In triangle $ABC$,if $a=13, b=8, c=7$,then $\cos(B+C) = $

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