Prove Heron's Formula (Trigonometry)

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Homework Help Overview

The discussion revolves around proving Heron's Formula for the area of a triangle given its side lengths. The formula involves the semiperimeter and is expressed as K=\sqrt{s(s-a)(s-b)(s-c)}. Participants are exploring various approaches to derive this formula using trigonometric identities and relationships.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to relate the area of the triangle to trigonometric functions, specifically using the sine of an angle. There are mentions of squaring the area formula and applying the cosine law, leading to algebraic expressions. Some participants express confusion about the steps taken by others.

Discussion Status

The discussion is ongoing, with participants sharing their individual approaches and questioning each other's methods. There is no clear consensus yet, but some guidance is being offered through the exploration of different algebraic manipulations and trigonometric identities.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There are indications of differing interpretations of the problem setup and the use of trigonometric identities.

rocomath
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Homework Statement



If a, b, c are the lengths of the sides of a triangle, then the area K of the triangle is given by [tex]K=\sqrt{s(s-a)(s-b)(s-c)}[/tex], where [tex]s=\frac{1}{2}(a+b+c)[/tex]. The number s is called the semiperimeter. Prove Heron's Formula. Hint: Use the area formula [tex]K=\frac{1}{2}bc\sin\phi[/tex].

sinphi should be sinA. it wouldn't let me use sinA.

The Attempt at a Solution



Absolute torture if you ask me! I need help getting on the right track, any help is appreciated.

Ignore everything from the triangle and down, that's a different problem.

http://img206.imageshack.us/img206/697/53149485pi4.jpg​
 
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my way:
K=whatever
K^2=whatever without sqrt

K^2 = that sin theta area^2

and using identidy sin^2 = 1-cos^2 in above

and cosine law, and some simplication, you would eventually reach somewhere like
-a^4+6a^3+3a^2...

and now just expand that herione thing

lol >(evil smile)<
 
rootX said:
my way:
K=whatever
K^2=whatever without sqrt

K^2 = that sin theta area^2

and using identidy sin^2 = 1-cos^2 in above

and cosine law, and some simplication, you would eventually reach somewhere like
-a^4+6a^3+3a^2...

and now just expand that herione thing

lol >(evil smile)<
lol ... i don't follow!
 
you didn't square the equation in the starting, but I did..
 
rootX said:
you didn't square the equation in the starting, but I did..
ok let me try it that way.
 

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