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Triangle Approximation Derivation

  • #1

Homework Statement



Here is a drawing with all the needed variables:

http://i.imgur.com/192GI.jpg


Homework Equations





The Attempt at a Solution



I have been trying to figure out how this approximation is derived for some time now and have no progress to show for it. Any help in figuring out the steps would be greatly appreciated.
 

Answers and Replies

  • #2
The law of cosines says that

b^2 - a^2 = c^2 - 2ac * cos(B),

so,

b - a = -2ac * cos(B) / (b+a) + c^2 / (b+a).

If one says that a ~ b because c << a, then the above becomes

b - a ~ -c * cos(B) + c^2 / (2a),

which is close but does not match and is way off for B approaching zero degrees.
 
  • #3
631
0
The law of cosines says that

b^2 - a^2 = c^2 - 2ac * cos(B),

so,

b - a = -2ac * cos(B) / (b+a) + c^2 / (b+a).

If one says that a ~ b because c << a, then the above becomes

b - a ~ -c * cos(B) + c^2 / (2a),

which is close but does not match and is way off for B approaching zero degrees.
As I read the figure and understand your notation, B = [tex]\pi[/tex] - ([tex]\theta[/tex] + [tex]\gamma[/tex]), thus what you got equals what's on the figure.
 
  • #4
As I read the figure and understand your notation, B = [tex]\pi[/tex] - ([tex]\theta[/tex] + [tex]\gamma[/tex]), thus what you got equals what's on the figure.
Almost but not quite. I cannot figure out where the sin^2(B) factor comes from.
 
  • #5
631
0
Are you familiar with Taylor series? I could get the form they have given using a Taylor series approximation.
 

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