In $a △ A B C, if a : b : c = 7 : 8 : 9$ then find $\cos A : \cos B : \cos C$

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Detailed Solution

Given $a : b : c = 7 : 8 : 9$

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{64 k^{2} + 81 k^{2} - 49 k^{2}}{2 (8 k) (9 k)}$

$= \frac{96 k^{2}}{144 k^{2}} = \frac{2}{3}$

$\cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a} = \frac{81 k^{2} + 49 k^{2} - 64 k^{2}}{2 (9 k) (7 k)}$

$= \frac{66 k^{2}}{126 k^{2}} = \frac{11}{21}$

$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{49 k^{2} + 64 k^{2} - 81 k^{2}}{2 (7 k) (8 k)}$

$= \frac{32 k^{2}}{112 k^{2}} = \frac{2}{7}$

$\cos A : \cos B : \cos C$

$= \frac{2}{3} : \frac{11}{21} : \frac{2}{7} = \frac{2}{3} \times 21 : \frac{11}{21} \times 21 : \frac{2}{7} \times 21 = 14 : 11 : 6$

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