I don't know a lot about the Heron formula.
But I know, the area of any given triangle is base*height divided by 2 (A= 1/2 * g*h).

If I have a triangle A, B, C, with the two sides ("a" and "b") given instead of base (="c") and height (="h"), and the angle between these sides (GAMMA), then the formula for computing the area is A = 1/2*a*b*sin(GAMMA).

And for example: if b is not given, but one of the other angles, say ALPHA instead, you can find out BETA (= 180° - GAMMA - ALPHA), and can use the sinus formula to find out "b":
a/sin(ALPHA) = b/sin(BETA) ----> b = a*sin(BETA)/sin(ALPHA)

So, the area A will be: A = (a^2)*sin(ALPHA)*sin(GAMMA)/(2*sin(BETA))

Is this okay, or is it too low level?

Unless I am reading it too quickly, your first situation, which we can call Side-Angle-Side, is correct. The "level" is fine, unless you want to look back at it and perhaps provide a different, more efficient or more elegant solution. (I make no claims as to whether such another solution exists.)
As for the second, I think we need to consider whether the side is included between the angles (Angle-Side-Angle) or not (Angle-Angle-Side). If you think such distinctions are not needed, please explain why and I will try to read more slowly next time. :)

okay, point taken, and accepted. Well, it's too long time ago that I messed around with such things, and I thought the symbols etc were the same everywhere. Oh, Just today i learned, that - in opposite to the German formulas - in the English language countries area has the symbol S ... sorry, my fault

The triangle ABC I meant was:
the angle [ALPHA] is on point A, between the sides b and c, the angle [BETA] on point B, between the sides c and a, and the angle [GAMMA] on point C, between the sides a and b.

So, while the normal formula S=g*h/2 is used for a known base g (in my triangle it would be c=g), and a known height h, my first solution was for Side-Angle-Side, the second one (with just side "a" known, and [ALPHA] and [GAMMA] Angle-Angle-Side: S=(a^2)*sin(ALPHA)*sin(GAMMA)/(2*sin(BETA)).

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Thorin:

Well, the issue wasn't the label for area. We're used to A for area in the U.S.
What I was trying to clarify was whether you were investigating Angle-Angle-Side or Angle-Side-Angle and I guess you were considering the latter.

I get the same thing you do for Angle-Side-Angle, but I would re-write alpha as 180-(beta+gamma) so that it is only expressed in terms of the assumed measurements.
If you worked this out yourself, then I am pleased that you used your solution for Angle-Side-Angle to solve Angle-Angle-Side, rather than go back to putting the A-A-S question in terms of A = (1/2)bh (area of a triangle formula w/ the usual letters used here).
So, what you did was called "folding back," meaning you let your work on the first become a piece of core knowledge rather than going back to only what had been previously your core knowledge.

Also, I was hoping outside the U.S. that more people knew about Heron's Formula for computing the area of a triangle just from knowing the lengths of the sides.
See if you can develop it, and then see my posts on medians, angle bisectors and altitudes. :)

Or any of the others I posted besides the area of a triangle question.

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@Sarge: After learned of this Huron formula just today, I give up to prove it. I fear I either forgot it entirely, or I never had it when I was in school.

But I'm willing to take the next challenge:

Two poles are perpendicular to the ground, set a certain distance apart. The top of each pole is connected to the ground by a wire. The two wires meet on the ground somewhere in between the two poles. Where should the point of meeting be in order to minimize the amount of wire used? (You may assume the ground between the two poles is a line segment.)

Question: These poles, are they supposed to have the same height?

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@Sarge: After learned of this Huron formula just today, I give up to prove it. I fear I either forgot it entirely, or I never had it when I was in school.

But I'm willing to take the next challenge:

Two poles are perpendicular to the ground, set a certain distance apart. The top of each pole is connected to the ground by a wire. The two wires meet on the ground somewhere in between the two poles. Where should the point of meeting be in order to minimize the amount of wire used? (You may assume the ground between the two poles is a line segment.)

Question: These poles, are they supposed to have the same height?

Good question, Thorin.
The answer is, you get to choose.
See Georg Polya's (1945) "How to Solve It."

Regarding Heron's Formula:

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oh...now that makes perfect sense...Thanks for elucidating that for us.

:-)

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Test post: [font=symbol]a[/font]

Edit: Sweet, it worked the way in which I expected!

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You all seem like clever people.

So can someone please, please, explain what is happening here.

Relieve me of this torture.

Well Graeme, have you got the hang of this now?

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Awaiting the return of Thorin and Jason Gee ... .

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Awaiting the return of Thorin and Jason Gee ... .

LOL, I'm here, 6am with no sleep again, perfect for some geometrical gymnastics :P, waddup? more triangles?

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"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Awaiting the return of Thorin and Jason Gee ... .

LOL, I'm here, 6am with no sleep again, perfect for some geometrical gymnastics :P, waddup? more triangles?

See any of the following: Problem 1; Problem 2; Problem 3; Problem 5; or Problem 6. You and Thorin have already been considering Problem 4.

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Awaiting the return of Thorin ... .

Today I was a bit too busy with problems of my desktop. But I think tomorrow or Monday I have time and mood enough to take the challenge :D
I haven't forgotten it yet

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Working on Problem 2 (because I like wires & poles)

notes so far:
- poles can be different heights
- smallest length of wire is always where, angle between wire segment #1 and ground is equal to angle between wire segment #2 and ground.
- can prove with pythag + a little calc

gotta sleep :D

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"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Well Graeme, have you got the hang of this now?

LOL. Maybe in the next life :-D

Working on Problem 2 (because I like wires & poles)

notes so far:
- poles can be different heights
- smallest length of wire is always where, angle between wire segment #1 and ground is equal to angle between wire segment #2 and ground.
- can prove with pythag + a little calc

gotta sleep :D

Hmmm. What's this angles thingy?

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Capitalize on this good fortune, one word can bring you round ... changes.

Working on Problem 2 (because I like wires & poles)

notes so far:
- poles can be different heights
- smallest length of wire is always where, angle between wire segment #1 and ground is equal to angle between wire segment #2 and ground.
- can prove with pythag + a little calc

gotta sleep :D

Hmmm. What's this angles thingy?

I think it has something to do with triangles being 'pointy'.

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I did NOT authorize this belly writing!

jason gee wrote:

notes so far:
- poles can be different heights
- smallest length of wire is always where, angle between wire segment #1 and ground is equal to angle between wire segment #2 and ground.

I tend to second that.

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Determine which is "larger": sqrt(1+x^2)+sqrt(1+y^2) or sqrt(4+(x+y)^2).
Once you have solved it, state a related more general problem. Solve it and include your proof.

As far as I can tell it, it's the same result. I'm just working on the proof why it is the same :D

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jason gee wrote:

notes so far:
- poles can be different heights
- smallest length of wire is always where, angle between wire segment #1 and ground is equal to angle between wire segment #2 and ground.

I tend to second that.

Sadly I won't get a chance to write a proof until next weekend, I'm back at school finishing some exams on Transistors. If somebody's calculus isn't as rusty as mine they might have a crack at the proof:

it should be close to:
- Call the line segment along the ground L, the distance between the two poles
- Call the base of 1st wire's triangle X ( along L but shorter )
- other triangle's base is L-X
1) define the wirelengths as a functions of their base(X or L-X) & height ( pythagoras )
2) add the two functions to get a function for total wire length ( with respect to L, X, poleheight1 & poleheight2 )
3) take the first derivative of that function
4) Set the first derivative to 0 ( to find a relative extrema )
5) used the solved functions as argunents to TAN functions on both sides to show the angles are equal on both sides.

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"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

it should be close to:
- Call the line segment along the ground L, the distance between the two poles
- Call the base of 1st wire's triangle X ( along L but shorter )
- other triangle's base is L-X
1) define the wirelengths as a functions of their base(X or L-X) & height ( pythagoras )
2) add the two functions to get a function for total wire length ( with respect to L, X, poleheight1 & poleheight2 )
3) take the first derivative of that function
4) Set the first derivative to 0 ( to find a relative extrema )
5) used the solved functions as argunents to TAN functions on both sides to show the angles are equal on both sides.

Derivative with respect to what as the variable?
Keep at it.

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When I get around to going into the geometry/shapes manufacturing business you people are first on the 'to hire' list. I'll be paying top dollar.

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Founder of BOINC team Objectivists. Oh the humanity! Rational people crunching data!
I did NOT authorize this belly writing!