In a triangle ABC with usual notations,if a:b | vedclass

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(A) Given $a=7k, b=8k, c=9k$. Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{(8k)^2+(9k)^2-(7k)^2}{2(8k)(9k)} = \frac{64+81-49}{144}

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$14:11:6$

Vedclass · Answer

(A) Given $a=7k, b=8k, c=9k$. Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{(8k)^2+(9k)^2-(7k)^2}{2(8k)(9k)} = \frac{64+81-49}{144} = \frac{96}{144} = \frac{2}{3}$. Similarly,$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{(7k)^2+(9k)^2-(8k)^2}{2(7k)(9k)} = \frac{49+81-64}{126} = \frac{66}{126} = \frac{11}{21}$. And $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{(7k)^2+(8k)^2-(9k)^2}{2(7k)(8k)} = \frac{49+64-81}{112} = \frac{32}{112} = \frac{2}{7}$. Now,$\cos A : \cos B : \cos C = \frac{2}{3} : \frac{11}{21} : \frac{2}{7}$. Multiplying by $21$,we get $14 : 11 : 6$.

In a triangle $ABC$ with usual notations,if $a:b:c = 7:8:9$,then $\cos A : \cos B : \cos C =$

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