ratios and trig ratios

**earthboy** · November 7th 2011, 04:47 AM

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.

please help me...

**earboth** · November 7th 2011, 07:00 AM

Originally Posted by earthboy

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

please help me...

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:

$\cos(\alpha) : \cos(\beta) : \cos(\gamma) = 14 : 11 : 6$

**earthboy** · November 7th 2011, 09:45 AM

yes there's a little confusion with the question. i have edited it

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