In triangle ABC,if a: b: c = 3: 5: 7,then cos | vedclass

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(C) Given the ratio of sides as $a: b: c = 3: 5: 7$,let $a = 3x, b = 5x, c = 7x$.Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\co

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$\frac{12}{7}$

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(C) Given the ratio of sides as $a: b: c = 3: 5: 7$,let $a = 3x, b = 5x, c = 7x$.Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Substituting the values:$\cos A = \frac{(5x)^2 + (7x)^2 - (3x)^2}{2(5x)(7x)} = \frac{25x^2 + 49x^2 - 9x^2}{70x^2} = \frac{65x^2}{70x^2} = \frac{13}{14}$.$\cos B = \frac{(3x)^2 + (7x)^2 - (5x)^2}{2(3x)(7x)} = \frac{9x^2 + 49x^2 - 25x^2}{42x^2} = \frac{33x^2}{42x^2} = \frac{11}{14}$.Therefore,$\cos A + \cos B = \frac{13}{14} + \frac{11}{14} = \frac{24}{14} = \frac{12}{7}$.

In $\triangle ABC$,if $a: b: c = 3: 5: 7$,then $\cos A + \cos B =$

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