I've always thought you cannot add up denominators

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[PLAIN]http://img546.imageshack.us/img546/9343/math1234.jpg

Is this right?
I've always thought you cannot add up denominators...
 
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Let a/sin55=b/sin80=c/sin45=D
then a=Dsin55, b=Dsin80, c=Dsin45
(a+b+c)/(sin55+sin80+sin45)
=(Dsin55+Dsin80+Dsin45)/(sin55+sin80+sin45)
=D(sin55+sin80+sin45)/(sin55+sin80+sin45)
=D
 
Welcome to Physics Forums, nevmx.

That sum is not correct.

Is that all you need, or are you looking for a reason?
 
Some Pig said:
Let a/sin55=b/sin80=c/sin45=D
then a=Dsin55, b=Dsin80, c=Dsin45
(a+b+c)/(sin55+sin80+sin45)
=(Dsin55+Dsin80+Dsin45)/(sin55+sin80+sin45)
=D(sin55+sin80+sin45)/(sin55+sin80+sin45)
=D

I think I stand corrected: the algebra is compelling, but I am not seeing what is going on. It looks like some geometric relationship is setting a constraint that makes it possible.

The denominator is 2.511. I think what's confusing me is that there are an number of values for a, b and c given those angles...
 
nevmx, it is wrong to replace some fractions added together with a signal fraction by adding the the numerators together and dividing that by the sum of the denominators.

If some fractions are equal to each other that is a different situation, though. Apparently, these fractions are also all equal to the sum of the numerators divided by the sum of the denominators. This can be shown using a technique like Some Pig showed.
 
I think I see why now. This is saying that when you add three equal ratios together, the ratio is preserved in the sum.
1/2 + 2/4 + 5/10 = 8/16

Pretty cool.
 
welcome to pf!

hi nevmx! welcome to pf! :smile:
nevmx said:
Is this right?

short answer: yes :biggrin:

longer answer: if the ratio is 1/r, it's (a + b + c)/(ra + rb + rc) :wink: