With usual notations,in triangle ABC,if a=2, | vedclass

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(D) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.Substituting th

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$12$

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(D) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.Substituting these into the given expression:$\frac{b^2+c^2-a^2}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc} = \frac{k+7}{30}$$\frac{a^2+b^2+c^2}{2abc} = \frac{k+7}{30}$Given $a=2, b=3, c=5$,we have $a^2+b^2+c^2 = 4+9+25 = 38$ and $2abc = 2 	imes 2 	imes 3 	imes 5 = 60$.So,$\frac{38}{60} = \frac{k+7}{30}$$\frac{38}{60} = \frac{2(k+7)}{60}$$38 = 2k + 14$$2k = 24 \Rightarrow k = 12$.

With usual notations,in $\triangle ABC$,if $a=2, b=3, c=5$ and $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{k+7}{30}$,then $k=$

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