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Study Notes 035:
Trigonometry 9:
By William E. Steinman:
There is one other case of solving for the area of a triangle.
This is a hairy one where we are given the three sides of a triangle and no angles.
Let’s give it a go.
It is important to follow this through step-by-step so you really get it.
Make sure you have a few sheets of scratch paper.
Here is our favorite triangle again.
This formula is known as Heron’s formula. Heron, or Hero is the scientist who derived it.
To prove this formula, we can first recall the law of cosines.
a2 = b2 + c2 – 2bc cos A
or, cos A = b2 + c2 – a2/2bc
So 1 – cos A = 1 – (b2 + c2 – a2/2bc)
We can do some algebraic manipulation of this to arrive at this:
1 – cos A = ((a – b + c)(a+ b – c))/2bc
Similarly 1 + cos A = 1 + (b2 + c2 – a2/2bc)
And this will eventually yield 1 + cos A = ((b + c + a)(b + c + -a))/2bc
As I said, tedious.
We know that a + b + c = 2s. That is how we defined it.
And a – b + c = (a + b + c) -2b
All we did here was to add b and subtract b.
We can substitute and get 2s – 2b = 2(s - b)
With similar manipulation we can get
a + b – c = 2(s – c)
and b + c – a = 2(s – a)
Now we can recall a couple of our half angle identities.
Stated differently we can get
Sin2 ½ A = ½ (1 – cos A)
Substituting from the above stuff this becomes ((s – b)(s – c))/bc
Cos2 ½ A = ½ (1 + cos A) which yields s(s – a)/bc
We should notice that these are only true when A is less than 900 so all functions are positive.
Given all of that, we can now cut to the chase.
Do you have a headache?
Me too!
Imagine how Heron felt. He did this from scratch.
Later, after you have rested, you may want to go through this again to make sure you have it.
Next time we can get into the inverse trig functions.