If $\displaystyle a:b:c=7:8:9 $ ,

find $\displaystyle \cos A:\cos B:\cos C $

where a,b,c are the sides opposite to angles A,B,C in triangle ABC.

please help me...

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- Nov 7th 2011, 04:47 AMearthboyratios and trig ratios
If $\displaystyle a:b:c=7:8:9 $ ,

find $\displaystyle \cos A:\cos B:\cos C $

where a,b,c are the sides opposite to angles A,B,C in triangle ABC.

please help me... - Nov 7th 2011, 07:00 AMearbothRe: ratios and trig ratios
1. I assume that the a in cos(a) denotes the angle opposite the side a. To avoid some confusion I'll use $\displaystyle \alpha$ to denote this angle. Consequently $\displaystyle \beta, \gamma$ denote the angles opposite the sides b or c.

2. Use a triangle with the side lengthes

a = 7, b = 8, c = 9

3. Use the Cosine rule to determine the cosine values of the corresponding angles.

I've got:

$\displaystyle \cos(\alpha) : \cos(\beta) : \cos(\gamma) = 14 : 11 : 6$ - Nov 7th 2011, 09:45 AMearthboyRe: ratios and trig ratios
yes there's a little confusion with the question. i have edited it