# ratios and trig ratios

• November 7th 2011, 05:47 AM
earthboy
ratios and trig ratios
If $a:b:c=7:8:9$ ,
find $\cos A:\cos B:\cos C$
where a,b,c are the sides opposite to angles A,B,C in triangle ABC.

• November 7th 2011, 08:00 AM
earboth
Re: ratios and trig ratios
Quote:

Originally Posted by earthboy
If $a:b:c=7:8:9$ ,
find $\cos a:\cos b:\cos c$

1. I assume that the a in cos(a) denotes the angle opposite the side a. To avoid some confusion I'll use $\alpha$ to denote this angle. Consequently $\beta, \gamma$ 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:

$\cos(\alpha) : \cos(\beta) : \cos(\gamma) = 14 : 11 : 6$
• November 7th 2011, 10:45 AM
earthboy
Re: ratios and trig ratios
yes there's a little confusion with the question. i have edited it