In triangle ABC,if a: b: c = 4: 5: 6,then fra | vedclass

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(C) Given $a: b: c = 4: 5: 6$. Let $a = 4k, b = 5k, c = 6k$ for some constant $k > 0$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}

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$2$

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(C) Given $a: b: c = 4: 5: 6$. Let $a = 4k, b = 5k, c = 6k$ for some constant $k > 0$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5k)^2 + (6k)^2 - (4k)^2}{2(5k)(6k)} = \frac{25k^2 + 36k^2 - 16k^2}{60k^2} = \frac{45}{60} = \frac{3}{4}$. $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4k)^2 + (6k)^2 - (5k)^2}{2(4k)(6k)} = \frac{16k^2 + 36k^2 - 25k^2}{48k^2} = \frac{27}{48} = \frac{9}{16}$. $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4k)^2 + (5k)^2 - (6k)^2}{2(4k)(5k)} = \frac{16k^2 + 25k^2 - 36k^2}{40k^2} = \frac{5}{40} = \frac{1}{8}$. Now,calculate the expression: $\frac{\cos A + 3 \cos C}{\cos B} = \frac{\frac{3}{4} + 3(\frac{1}{8})}{\frac{9}{16}} = \frac{\frac{3}{4} + \frac{3}{8}}{\frac{9}{16}} = \frac{\frac{6+3}{8}}{\frac{9}{16}} = \frac{\frac{9}{8}}{\frac{9}{16}} = \frac{9}{8} 	imes \frac{16}{9} = 2$.

In $\triangle ABC$,if $a: b: c = 4: 5: 6$,then $\frac{\cos A + 3 \cos C}{\cos B} = $

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