How to find a:b:c given that s-a:s-b:s-c = 1:2:3

  • Thread starter Thread starter zorro
  • Start date Start date
  • Tags Tags
    Ratios
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
7 replies · 2K views
zorro
Messages
1,378
Reaction score
0

Homework Statement



Let s-a : s-b : s-c :: 1:2:3

then how do we find a:b:c from this?


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Try to format your expression in such a way that it is unambiguous what you are looking for.
 
Abdul Quadeer said:

Homework Statement



Let s-a : s-b : s-c :: 1:2:3

then how do we find a:b:c from this?
Rewrite this proportion as three equations. The proportion is saying is that s - b is 2 times s - a, s - c is 3 times s - a, and s - c is (3/2) times s - b.

That should give you somewhere to start.
 
Abdul Quadeer said:

Homework Statement



Let s-a : s-b : s-c :: 1:2:3

then how do we find a:b:c from this?


Homework Equations





The Attempt at a Solution

That's slightly confusing because it represents several proportions together.
You can analyze it as [itex]s-a: s-b::1:2[/itex], [itex]s-b: s-c::2: 3[/itex], and [itex]s-a: s-c::1: 3[itex]. Those can be written as fraction:<br /> [tex]\frac{s-a}{s-b}= \frac{1}{2}[/tex]<br /> [tex]\frac{s-b}{s-c}= \frac{2}{3}[/tex]<br /> and<br /> [tex]\frac{s-a}{s-c}= \frac{1}{3}<br /> <br /> The first equation can be rewritten as 2(s- a)= s- b, 3(s-b)= 2(s- c), and 3(s-a)= s- c.<br /> <br /> You can solve each of those for s: 2s- 2a= x- b so s= 2a- b. 3s- 3b= 2s- 2c so s= 3b- 2c. 3s- 3a= s- c so 2s= 3a- c or s= (3/2)a- (1/2)c. <br /> <br /> Now put them back together: s= 2a- b= (2/3)a- (1/2 c, s= 2a- b= 3b- 2c. You should be able to find the relationships between a, b, and c from that.[/tex][/itex][/itex]
 
HallsofIvy said:
That's slightly confusing because it represents several proportions together.
You can analyze it as [itex]s-a: s-b::1:2[/itex], [itex]s-b: s-c::2: 3[/itex], and [itex]s-a: s-c::1: 3[itex]. Those can be written as fraction:<br /> [tex]\frac{s-a}{s-b}= \frac{1}{2}[/tex]<br /> [tex]\frac{s-b}{s-c}= \frac{2}{3}[/tex]<br /> and<br /> [tex]\frac{s-a}{s-c}= \frac{1}{3}<br /> <br /> The first equation can be rewritten as 2(s- a)= s- b, 3(s-b)= 2(s- c), and 3(s-a)= s- c.<br /> <br /> You can solve each of those for s: 2s- 2a= x- b so s= 2a- b. 3s- 3b= 2s- 2c so s= 3b- 2c. 3s- 3a= s- c so 2s= 3a- c or s= (3/2)a- (1/2)c. <br /> <br /> Now put them back together: s= 2a- b= (2/3)a- (1/2 c, s= 2a- b= 3b- 2c. You should be able to find the relationships between a, b, and c from that.[/tex][/itex][/itex]
[itex][itex][tex] <br /> That made it more clear.<br /> I got a:b:c :: 5:4:3<br /> Thanks a lot Hallsofivy[/tex][/itex][/itex]