Triangle Approximation Derivation

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Homework Help Overview

The discussion revolves around the derivation of a triangle approximation using the law of cosines. Participants are examining the relationship between the sides of a triangle and the angles involved, particularly in the context of small angle approximations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate the law of cosines to derive an approximation, questioning the validity of their steps and assumptions, especially regarding the behavior of the angles as they approach zero. There is also mention of using Taylor series for approximation.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning specific factors in the derivation, such as the presence of the sin²(B) term. Some guidance on using Taylor series has been suggested, but no consensus has been reached.

Contextual Notes

There is a reference to a drawing that contains necessary variables, but the specifics of the drawing are not included in the discussion. Participants are also grappling with the implications of small angle approximations and the relationships between the triangle's sides and angles.

GroupTheory1
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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.
 
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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.
 
GroupTheory1 said:
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.
 
Sourabh N said:
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.
 
Are you familiar with Taylor series? I could get the form they have given using a Taylor series approximation.
 

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