Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three corners
of the triangle?

My Zeus tables explain how to set up a drilling rig for a given
diameter, but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!

TIA

Richard

Richard wrote:
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three corners
of the triangle?

My Zeus tables explain how to set up a drilling rig for a given
diameter, but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!

TIA

Richard



More thought and less haste is an obvious answer (with apologies).


http://www.ajdesigner.com/index_math.php


Richard

:-(

In message , Richard
writes
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three corners
of the triangle?

My Zeus tables explain how to set up a drilling rig for a given
diameter, but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!


You mean like typing

"diameter of a circle on an equilateral triangle"

into google ?

which yields various site such as this ...

http://mathworld.wolfram.com/EquilateralTriangle.html

Rocket science, eh ?

--
geoff

"geoff" wrote in message
...
In message , Richard
writes
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three corners of
the triangle?

My Zeus tables explain how to set up a drilling rig for a given diameter,
but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!


You mean like typing

"diameter of a circle on an equilateral triangle"

into google ?

which yields various site such as this ...

http://mathworld.wolfram.com/EquilateralTriangle.html

Rocket science, eh ?

plonk

circumscribed radius of an equilateral triangle of side L = L * 0.5574

For 185mm side, this is 107mm

In message , R D S
writes
"geoff" wrote in message
...
In message , Richard
writes
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three corners of
the triangle?

My Zeus tables explain how to set up a drilling rig for a given diameter,
but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!


You mean like typing

"diameter of a circle on an equilateral triangle"

into google ?

which yields various site such as this ...

http://mathworld.wolfram.com/EquilateralTriangle.html

Rocket science, eh ?

plonk

plonker ...

--
geoff

On 20 Jul, 14:47, Richard wrote:
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three corners
of the triangle?

My Zeus tables explain how to set up a drilling rig for a given
diameter, but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!

TIA

Richard

If the equilateral triangle is converted into three equal sized
triangles within the main one, using the centre of the circle as the
common point where all three converge, you get three triangles with
two 30deg angles and one of 120deg, and the lines from the side of the
circle to the centre is the radius of the circle.

Half one of these triangle, you get a right angled triangle with a
30deg and 60 deg angles. Using trig, the 185mm triangle edge now = 2
x the (A)djacent side of the triangle in relation to the 30deg angle.

Formula is therefore (185/2) / cos(30) = 106.80. As a generic
formula, r = L/2 / cos(30).

http://www.google.co.uk/search?hl=en...grees%29&meta=

I'm sure there is a simpler way, but that was an enjoyable challenge
to me as someone who hadn't done any trig in about 20 years!!

Matt

wrote in message
...
On 20 Jul, 14:47, Richard wrote:
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three corners
of the triangle?

My Zeus tables explain how to set up a drilling rig for a given
diameter, but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!

TIA

Richard

If the equilateral triangle is converted into three equal sized
triangles within the main one, using the centre of the circle as the
common point where all three converge, you get three triangles with
two 30deg angles and one of 120deg, and the lines from the side of the
circle to the centre is the radius of the circle.

Half one of these triangle, you get a right angled triangle with a
30deg and 60 deg angles. Using trig, the 185mm triangle edge now = 2
x the (A)djacent side of the triangle in relation to the 30deg angle.

Formula is therefore (185/2) / cos(30) = 106.80. As a generic
formula, r = L/2 / cos(30).


You beat me to it :-)

I think it can be done (proofed) with no cos or sin involved IMHO.

Adam

In an earlier contribution to this discussion,
ARWadworth wrote:

wrote in message
...
On 20 Jul, 14:47, Richard wrote:
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three
corners of the triangle?

My Zeus tables explain how to set up a drilling rig for a given
diameter, but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!

TIA

Richard

If the equilateral triangle is converted into three equal sized
triangles within the main one, using the centre of the circle as the
common point where all three converge, you get three triangles with
two 30deg angles and one of 120deg, and the lines from the side of
the circle to the centre is the radius of the circle.

Half one of these triangle, you get a right angled triangle with a
30deg and 60 deg angles. Using trig, the 185mm triangle edge now = 2
x the (A)djacent side of the triangle in relation to the 30deg angle.

Formula is therefore (185/2) / cos(30) = 106.80. As a generic
formula, r = L/2 / cos(30).


You beat me to it :-)

I think it can be done (proofed) with no cos or sin involved IMHO.

Adam

Yes it can. I did it using Pythagoras and interesting chords. It comes out
as r=L/sqrt(3) - which is the same thing anyway because cos(30)=sqrt(3)/2
--
Cheers,
Roger
______
Email address maintained for newsgroup use only, and not regularly
monitored.. Messages sent to it may not be read for several weeks.
PLEASE REPLY TO NEWSGROUP!

On 21 Jul, 23:13, "Roger Mills" wrote:
In an earlier contribution to this discussion,





ARWadworth *wrote:
wrote in message
....
On 20 Jul, 14:47, Richard wrote:
Given an equilateral triangle of side 185 mm, how do I calculate the
diameter or radius of the circle that intersects with the three
corners of the triangle?

My Zeus tables explain how to set up a drilling rig for a given
diameter, but not the reverse.

The formula would be helpful.

A link to a web site of such useful information would be wonderful!

TIA

Richard

If the equilateral triangle is converted into three equal sized
triangles within the main one, using the centre of the circle as the
common point where all three converge, you get three triangles with
two 30deg angles and one of 120deg, and the lines from the side of
the circle to the centre is the radius of the circle.

Half one of these triangle, you get a right angled triangle with a
30deg and 60 deg angles. *Using trig, the 185mm triangle edge now = 2
x the (A)djacent side of the triangle in relation to the 30deg angle.

Formula is therefore (185/2) / cos(30) = 106.80. *As a generic
formula, r = L/2 / cos(30).

You beat me to it :-)

I think it can be done (proofed) with no cos or sin involved IMHO.

Adam

Yes it can. I did it using Pythagoras and interesting chords. It comes out
as r=L/sqrt(3) - which is the same thing anyway because cos(30)=sqrt(3)/2
--
Cheers,
Roger
______
Email address maintained for newsgroup use only, and not regularly
monitored.. Messages sent to it may not be read for several weeks.
PLEASE REPLY TO NEWSGROUP!- Hide quoted text -

- Show quoted text -

My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras
and trig. Why does sqrt(3)/2 = cos(30deg)? I can see that it does, I
just can't see the Pythagoras on the way to getting there......

Is it because of the defined relationship in a triangle with 30 / 60 /
90 degree angles (making the sides = 106.8, 53.4, 92.5 in this case?)

I should have done A level maths (perhaps I will one day) - its
fascintating stuff!

Matt

In an earlier contribution to this discussion,
wrote:

On 21 Jul, 23:13, "Roger Mills" wrote:
In an earlier contribution to this discussion,




I think it can be done (proofed) with no cos or sin involved IMHO.

Adam

Yes it can. I did it using Pythagoras and interesting chords. It
comes out as r=L/sqrt(3) - which is the same thing anyway because
cos(30)=sqrt(3)/2 --
Cheers,
Roger


My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras
and trig. Why does sqrt(3)/2 = cos(30deg)? I can see that it does, I
just can't see the Pythagoras on the way to getting there......

Is it because of the defined relationship in a triangle with 30 / 60 /
90 degree angles (making the sides = 106.8, 53.4, 92.5 in this case?)

I should have done A level maths (perhaps I will one day) - its
fascintating stuff!

Matt

First off, a correction to my previous post. I've no idea what an
"interesting" chord is - I of course meant "intersecting chords"!

To answer your question. . .

Imagine an equilateral triangle with side length 2 units. Draw a
perpendicular from apex to base. You now have two 30/60/90 triangles with
the 30 degree angle at the top. The hypotenuse is 2 units long and the side
opposite the 30 degrees is 1 unit. Apply Pythagoras to that and (adjacent
side)^2 = 2^2 - 1^2 = 4 - 1 = 3
The length of the adjacent side is thus sqrt(3) units. The cosine of 30
degrees - adjacent side/hypotenuse is therefore sqrt(3)/2

QED!
--
Cheers,
Roger
______
Email address maintained for newsgroup use only, and not regularly
monitored.. Messages sent to it may not be read for several weeks.
PLEASE REPLY TO NEWSGROUP!

dave wrote:

You say QED - but is it? Maybe being pedantic here but is this a proof
as such? I mean what you have written is correct - but it's "just" a
particular case that happens to give the result expected. As I say,
not being pedantic but wonder if this really is a "proof"?

I'd say that if anything is missing it's that Roger didn't /prove/ that
the perpendicular from one corner of the equilateral triangle to the
opposite side actually bisects the angle and the opposite side. It's
obvious from the symmetry that it does, but can you prove it starting
only from Euclid's axioms?

I'm no mathematician but the definition of a cosine is, I believe, a
series.

Hmm... As usually taught you start with the simple right-angled
triangle definitions - cos = adjacent/hypotenuse, etc. - and then extend
that definition to allow angles outside the range 0 to 90 deg. Then the
power series expansions can be derived and the trig functions
generalised to allow complex number arguments. Hence if you can prove
the bisection of the angle and side above and accept Pythagoras (which
can be proved in several ways) you know that the 1 : root 3 : 2 triangle
has angles of 30, 60 & 90 deg. and that cos 30 deg. = sqrt(3)/2, and so on.

This was second form stuff when I was at school. It's a bit worrying to
read things like

My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras
and trig. Why does sqrt(3)/2 = cos(30deg)?

(Discuss.)

--
Andy

On 23 Jul, 08:39, Andy Wade wrote:
dave wrote:
You say QED - but is it? Maybe being pedantic here but is this a proof
as such? I mean what you have written is correct - but it's "just" a
particular case that happens to give the result expected. As I say,
not being pedantic but wonder if this really is a "proof"?

I'd say that if anything is missing it's that Roger didn't /prove/ that
the perpendicular from one corner of the equilateral triangle to the
opposite side actually bisects the angle and the opposite side. *It's
obvious from the symmetry that it does, but can you prove it starting
only from Euclid's axioms?

I'm no mathematician but the definition of a cosine *is, I believe, a
series.

Hmm... *As usually taught you start with the simple right-angled
triangle definitions - cos = adjacent/hypotenuse, etc. - and then extend
that definition to allow angles outside the range 0 to 90 deg. *Then the
power series expansions can be derived and the trig functions
generalised to allow complex number arguments. *Hence if you can prove
the bisection of the angle and side above and accept Pythagoras (which
can be proved in several ways) you know that the 1 : root 3 : 2 triangle
has angles of 30, 60 & 90 deg. and that cos 30 deg. = sqrt(3)/2, *and so on.

This was second form stuff when I was at school. *It's a bit worrying to
read things like

My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras
and trig. *Why does sqrt(3)/2 = cos(30deg)?

(Discuss.)

--
Andy

Perhaps only worrying in that its now 20 years ago and (despite being
an accountant) I left maths far behind me so it may simply be my
forgetfulness.

Though I'm fairly sure that Pythagoras and trig were taught as
sequential steps - first learn Pythagoras such that the dimensions of
any right angled triangle can be worked out. Then learn trig in order
that angles or lengths can be derived based on given information.

So whilst we all knew SOHCAHTOA etc, and we all knew Asqr = Bsqr +
Csqr, the two formula were not merged into the (entirely logical as I
see it now) position above.

That's not to say, though, that with a bit of thought and a simple
explanation above I can't see how that works.

Perhaps what I should have written was "my 1998 Maths GCSE didn't
stretch to being explicit about formulae which merged Pyth and trig,
and instead kept them separate (but highly related of course) to make
a two step process rather than a single step"

Matt

wrote:
On 23 Jul, 08:39, Andy Wade wrote:

Hence if you can prove
the bisection of the angle and side above and accept Pythagoras (which
can be proved in several ways)

I think that is one of the few mathematical proofs I ever mastered! ;-)

This was second form stuff when I was at school. It's a bit worrying to
read things like

My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras
and trig. Why does sqrt(3)/2 = cos(30deg)?
(Discuss.)

I did GCE O level maths in '83 or '84, and don't recall trig getting
much beyond basic right angle triangle stuff, with a possible
requirement to be able to use the sine or cosine rules. I was not
sufficiently a fan of maths (at the time) to take it at A level.

I do remember having quite a nasty shock when doing maths at university
to find there was a whole world about trig identities and equivalents
that I knew absolutely nothing about! (Much of the difficulty stemming
from the fact that there were several maths groups one could be in, and
I happened to get the one where 95% of the students who had done A level
maths, and the very obtuse Welshman teaching, thought he could race
through it all as a "quick bit of revision"). After a few weeks of that
I decided to switch groups and found that the alternative was a running
at a much more sensible pace (the first group had almost finished course
in the first four weeks!)

Perhaps only worrying in that its now 20 years ago and (despite being
an accountant) I left maths far behind me so it may simply be my
forgetfulness.

It surprising really how even after having spent almost all of my
working life in quite technical roles, just how little maths one is
called upon to use.

Though I'm fairly sure that Pythagoras and trig were taught as
sequential steps - first learn Pythagoras such that the dimensions of
any right angled triangle can be worked out. Then learn trig in order
that angles or lengths can be derived based on given information.

So whilst we all knew SOHCAHTOA etc, and we all knew Asqr = Bsqr +

Hmm - yes remember that one ;-)

(I think I preferred one of my maths teachers mnemonics of "Two Old
Aunts Sat On High Chairs and Howled)

Perhaps what I should have written was "my 1998 Maths GCSE didn't
stretch to being explicit about formulae which merged Pyth and trig,
and instead kept them separate (but highly related of course) to make
a two step process rather than a single step"

That (I discovered later in life) was one of the problems I often had
with maths at that level - it was never really built from first
principles and hence never really satisfied my desire to know *why*
something worked.

A good example would be something like matrix operations. At O level
there was never any worthwhile application given for why you might want
to carry out these manipulations - which made them seem all rather
pointless. Its only later when you realise you can convolve data sets,
solve simultaneous equations, and do all sorts of fancy graphics with
them (to name but a few applications) that you realise they do actually
have a purpose.




--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

John Rumm coughed up some electrons that declared:


I did GCE O level maths in '83 or '84, and don't recall trig getting
much beyond basic right angle triangle stuff, with a possible
requirement to be able to use the sine or cosine rules. I was not
sufficiently a fan of maths (at the time) to take it at A level.

The AO-level maths did cover a bit more of this sort of thing, also in 83
for me. A level maths covered a lot more and definately was the level
physics at uni assumed.

I do remember having quite a nasty shock when doing maths at university
to find there was a whole world about trig identities and equivalents
that I knew absolutely nothing about! (Much of the difficulty stemming
from the fact that there were several maths groups one could be in, and
I happened to get the one where 95% of the students who had done A level
maths, and the very obtuse Welshman teaching, thought he could race
through it all as a "quick bit of revision"). After a few weeks of that
I decided to switch groups and found that the alternative was a running
at a much more sensible pace (the first group had almost finished course
in the first four weeks!)

Wave mechanics and associated maths blew my mind - never did get the hand of
that stuff.

Never used any of it since, being a sysadmin and programmer.

I fused a braincell trying to do 3d-trig on my roof last month. Fortuanately
the basics were still there but I had to work lots of things out the hard
way, starting from the beginning.

Cheers

Tim

In article ,
says...

So whilst we all knew SOHCAHTOA etc, and we all knew Asqr = Bsqr +

Hmm - yes remember that one ;-)

(I think I preferred one of my maths teachers mnemonics of "Two Old
Aunts Sat On High Chairs and Howled)

Silly Old Horse, Curley And Heavy, Trod On Albert

and " X equals minus B, plus or minus the square root of B squared minus
4 A C all over 2A" - Not used that in a very long time.

There's nothing wrong with learning by rote. I can still multipy by
reciting the tables, where necessary :-)
--
John W
To mail me replace the obvious with co.uk twice

"John Rumm" wrote in message
et...
wrote:
On 23 Jul, 08:39, Andy Wade wrote:

Hence if you can prove
the bisection of the angle and side above and accept Pythagoras (which
can be proved in several ways)

I think that is one of the few mathematical proofs I ever mastered! ;-)

This was second form stuff when I was at school. It's a bit worrying to
read things like

My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras
and trig. Why does sqrt(3)/2 = cos(30deg)?
(Discuss.)

I did GCE O level maths in '83 or '84, and don't recall trig getting much
beyond basic right angle triangle stuff, with a possible requirement to be
able to use the sine or cosine rules. I was not sufficiently a fan of
maths (at the time) to take it at A level.


I was lucky in that I enjoyed doing my GCE O level and did go on to do an A
level. I bumped into my A level teacher 20 years after doing my A level and
took the time to tell her she was the best teacher I ever had (she was no
longer teaching and was interviewing my wife for a job when I bumped into
her). I have just dragged some of my old schoolbooks out of the loft to help
a friends lad out who is starting A level next term.

She used to do maths things that were not on the course, just for fun.
Learning that any recurring decimal can be written as a fraction is my
favourite

eg .123123123 recurring or any other numbers (I have used 3 digits here but
you can use more eg .1234512345) will always be a fraction if

(time to use letters for numbers and remember that these are recurring
decimals)

..abcabcabc will be the fraction abc/999
..abcdabcdabcd will will be the fraction abcd/9999.

Adam

In message , ARWadworth
writes
I did GCE O level maths in '83 or '84, and don't recall trig getting
much beyond basic right angle triangle stuff, with a possible
requirement to be able to use the sine or cosine rules. I was not
sufficiently a fan of maths (at the time) to take it at A level.


I was lucky in that I enjoyed doing my GCE O level and did go on to do
an A level. I bumped into my A level teacher 20 years after doing my A
level and took the time to tell her she was the best teacher I ever had
(she was no longer teaching and was interviewing my wife for a job when
I bumped into her). I have just dragged some of my old schoolbooks out
of the loft to help a friends lad out who is starting A level next term.




by far my best teacher at "O" level was "Nut" Hayes, who was B.Sc.
Physics (Failed), stirred his tea with his comb etc ... but was prolly
the best teacher I came across

Respect ... as one would say now



--
geoff