Is it an extradimensional monster or just one of the plain variety?

I suppose he could be trapped [ In the Phantom Zone ] like the three convicted war criminals in the first superman movie, he seems to be making the same expression .... [ Oh no, See what happens when you discuss trigonomentry at 4:30 am ? ]

---

"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.

Try it. Provide details.

Okay, give a [ new ] example [or repeat one if I missed it] and I'll solve it, It wouldn't be fair if I make it myself.

What about two side and a non-included triangle?

You mean leave all angles unknown? nope won't work unless you know the angle between them. can do it if it's a right triangle of course( because you then know one angle, the known sides have to be the short sides in this case & use pythagorus, more work might reveal other side, have to see example to tell)
[Edit: cancel that, I see what you mean now, given not included angle? then use the Law of sines, which can solve for certain cases]

For the other cases you've mentioned, are the right triangle definition of the three basic trigonometric functions all that you will need? What other trigonometry might be used?

naturally you can use pythagoras' theorem to shortcut in a couple of places to avoid inverse trig functions.
[Edit: also law of sines & law of cosines , if still stuck]

Yes, the Law of Sines and the Law of Cosines (generalized Pythagorean Theorem) will be quite helpful.
If all angles are unknown and all three sides measures ARE known, it can still be done.

What if you know the lengths of all three medians?
What if you know the lengths of all three altitudes (even if some of the altitudes go outside the triangle)?
What if you know the lengths of the bisectors of the interior angles (assume by this, though angle bisectors are rays, that I mean the length of the sub-segments of these rays that are not exterior to the triangle)?

---

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

Getting more complicated now ...

What if you know the lengths of all three medians?

I know there's a formula relating area to medians, but I'd have to look that one up...so it's reasonable that lengths etc.. could be derived. I remember another relationship among the medians themselves bisect each other at 2/3 the others length, that could be useful.

What if you know the lengths of all three altitudes (even if some of the altitudes go outside the triangle)?

should that form a rotated copy of the triangle?
[Edit: Nope :(, oh well have to look that one up too) "one half of the product of an altitude's length and its base's length equals the triangle's area." might be a start]

What if you know the lengths of the bisectors of the interior angles (assume by this, though angle bisectors are rays, that I mean the length of the sub-segments of these rays that are not exterior to the triangle)?

isn't that medians again?
[Edit: quick lookup yields "Trigonometric functions of half angles in a triangle can be expressed in terms of the triangle sides as ... " followed by Many functions of triangle sides + semiperimeter. :S ]

[Later: come to think of it, shouldn't enough components geometrically derived from a triangle lead back to the original triangle?]

---

"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.

And does it still add up if there is a monster inside ?

Yes... ;)

---

It may not be 1984 but George Orwell sure did see the future . . .

And does it still add up if there is a monster inside ?

Yes... ;)

HeHeHe. Even I can understand this one:))

---

Pure mathematics is, in its way, the poetry of logical ideas.

Albert Einstein

Someone's created a monster. A friendly, mathematically defined monster.

---

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

http://mathworld.wolfram.com/Monster-Barring.html

---

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

http://mathworld.wolfram.com/Monster-Barring.html

Thanks for the link. I'll coopt that term to describe the concept to particularly slippery polemicsts when arguing epistemology and ethics. Much appreciated.

---

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

Getting more complicated now ...

What if you know the lengths of all three medians?

I know there's a formula relating area to medians, but I'd have to look that one up...so it's reasonable that lengths etc.. could be derived. I remember another relationship among the medians themselves bisect each other at 2/3 the others length, that could be useful.

Try to do it without looking it up. You could, however, double-check the info about 2/3. You are correct, but need to be more specific. Another way of putting it is the two sub-segments of the medians are in a 2:1 length ratio. The specificity is needed regarding which sub-segment is twice as long as the other?
If you keep with algebra and trigonometry only, you'll wind up with a messy set of simultaneous equations. If I recall correctly, it's a 3 by 3 system, but quadratic not linear.
There's a near geometrical approach to this instead.
Overall, the last three I posed tend to go in the same direction: messy if one maintains an algebraic/trigonometric outlook, rather than seeking out additional geometric relations.

---

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

Fair enough! a rough sketch made it much easier. add up bases of inner right triangles created by the medians ( calculated from known 2/3 & 1/3 medians, using pythagoras' ), LOL avoiding trig where possible is a good thing :D

---

"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.

Fair enough! a rough sketch made it much easier. add up bases of inner right triangles created by the medians ( calculated from known 2/3 & 1/3 medians, using pythagoras' ), LOL avoiding trig where possible is a good thing :D

From the sounds of it, you may have not made your sketch look general enough and it led you to assume something you shouldn't. (Euclid was also guilty of this in a small number of cases.)

If your sketch looked isosceles or equilateral, then some or all medians will be also altitudes. In the general case, medians are, of course, something altogether different. Thus, the sub-triangles are not going to be right triangles in general.

---

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

Fair enough! a rough sketch made it much easier. add up bases of inner right triangles created by the medians ( calculated from known 2/3 & 1/3 medians, using pythagoras' ), LOL avoiding trig where possible is a good thing :D

From the sounds of it, you may have not made your sketch look general enough and it led you to assume something you shouldn't. (Euclid was also guilty of this in a small number of cases.)

If your sketch looked isosceles or equilateral, then some or all medians will be also altitudes. In the general case, medians are, of course, something altogether different. Thus, the sub-triangles are not going to be right triangles in general.

All true, good spot! well looking closer....

---

"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.

All true, good spot! well looking closer....

Alright, I think , should get area of triangle, from (area of) median triangle * 4/3. six inner triangles appear equal in area. so use known 1/3 & 2/3 median lengths and one sixth total area to determine ... mental block.... :S .... Ooh treat as a base & height of a right triangle and use pythag to determine third side ... then add em up (six of them )

any closer ? appears to work (on my dodgy sketches )

---

"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.

...Doh, use the triangle area from sides (rearranged) for that bit, not pythag+mofified triangle.

---

"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.

All true, good spot! well looking closer....

Alright, I think , should get area of triangle, from (area of) median triangle * 4/3. six inner triangles appear equal in area. so use known 1/3 & 2/3 median lengths and one sixth total area to determine ... mental block.... :S .... Ooh treat as a base & height of a right triangle and use pythag to determine third side ... then add em up (six of them )

any closer ? appears to work (on my dodgy sketches )

If the median triangle is defined as the triangle with side lengths equal to the lengths of the medians of the original triangle, then, yes, if memory serves me correctly the area of the original triangle = (4/3) * the area of the median triangle. So, just apply Heron's Formula to get the area of the median triangle and multiply by 4/3.

The things to check, then, are that you actually get a unique triangle (up to isomorphism) when talking about a median triangle and that area of the original triangle = (4/3) * the area of the median triangle.

I don't recall a comparison of the areas of the six inner triangles.

---

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

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?

---

Account frozen...

Try these two tests to win £500

---

---

Pure mathematics is, in its way, the poetry of logical ideas.

Albert Einstein

correct! You win, go to beijing collect 500 pounds. :D

---

"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.

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. :)

---

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