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

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

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$12: 9: 2$

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(A) Given the sides $a: b: c = 4: 5: 6$,let $a = 4k$,$b = 5k$,and $c = 6k$ for some constant $k > 0$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5k)^2 + (6k)^2 - (4k)^2}{2(5k)(6k)} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = \frac{3}{4}$. $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4k)^2 + (6k)^2 - (5k)^2}{2(4k)(6k)} = \frac{16 + 36 - 25}{48} = \frac{27}{48} = \frac{9}{16}$. $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4k)^2 + (5k)^2 - (6k)^2}{2(4k)(5k)} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = \frac{1}{8}$. Now,find the ratio $\cos A : \cos B : \cos C = \frac{3}{4} : \frac{9}{16} : \frac{1}{8}$. Multiply by the least common multiple of the denominators,which is $16$: $\cos A : \cos B : \cos C = (\frac{3}{4} 	imes 16) : (\frac{9}{16} 	imes 16) : (\frac{1}{8} 	imes 16) = 12 : 9 : 2$.

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

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