Geometry help needed

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I want to build a right triangle, 30 deg. 60 deg. angles without using a protractor (don't have one). The long arm will be 24 inches and will be using a square for the right angle. Anyone know the formula to get the length of the short side? I can figure out the hypotenuse for a double check of squareness.
Geometry is in my misty past by over 50 years.
--
Gerald Ross
Cochran, GA
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Gerald Ross wrote:

The short side is long side/2
The medium side is short side x square root of 3.
These only apply for 30-60-90 triangles.
For your situation, the shortest side is 24"/1.732 = 13.857
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On Sun, 09 Apr 2006 08:35:04 -0500, Dave Nay

That's possibly the way I'd do it, thanks to Pythagoras ...or go out and buy a protractor, or compasses. I prefer compasses [or a trammel] for specific angles like that [you can have, or make large ones and get better accuracy over distance].
Or, having a computer to be able to post here, there is CAD. Print a sheet with the angle, fold along the lines and use as a template.
Or there is a way to trisect an angle [approximate, but not that you'd notice] using the square. So a 30 degree angle is possible. All physical measures are inherently approximate even if not in theory, so it would do.
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On Sun, 09 Apr 2006 09:30:08 -0400, Gerald Ross wrote:

Short side is 1x, hypotenuse is 2x, and long side is 1.732x (sqroot of 3). So your short side is a little less than 13 7/8.
D.G. Adams
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On Sun, 09 Apr 2006 09:30:08 -0400, Gerald Ross

13.856"
That's okay, it's trigonometry you needed anyway :-)
Taking the 30 degree angle, find the tangent and multiply it by 24"
Without trig tables or a calculator that can do trig, tan(30) = 1/3 * sqrt(3) and everybody knows the sqrt(3) = 1.73, right?
Of if you want to consider the 60 degree angle first, the tan sqrt(3), then the short side = 24 / 1.73.
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wrote:

Why not use geometry? For a 30 angle you can lay it out more easily by geometry.
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"you can lay it out more easily by geometry."
Not sure what you mean, Andy.
Perhaps you would care to post your solution here?
My searching seemed to indicate it was a Trig problem. But, then, I could NOT find a solution worth posting. If you did, share.

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On Mon, 10 Apr 2006 03:16:56 +0000, Gooey TARBALLS opined:

Draw a long line (1). Construct a perpendicular (2). Set dividers to desired length of shortest leg (x). Strike that distance on the perpendicular, starting at the intersection (origin, o). Step off twice the divider setting on the first, horizontal line. Construct a perpendicular at the 2x point (3). Strike the divider distance up the second perpendicular (4). Connect the two upper points (5). (You now have a 1x2 rectangle.) Set the dividers to 2x. Strike an arc which intersects the upper horizontal (parens). Construct a third perpendicular to the horizontal line which intersects the intersection of the second horizontal and the 2x arc (6). Connect the origin to this last intersection (o, +). QED. 2 4 x .------------+--.-- 5 | ) | | |) | | | )| _o_______.____|__|__ 1 x 2x 6 3
--
"Keep your ass behind you"
wreck20051219 at spambob.net
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SNIP etc...
Depending on age, he may not have studied any Euclidian geometry at all. No proofs by construction in the last two "geometry" texts at our school.
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Draw a vertical line desired length of shortest leg (x). Draw a horizontal line 2x in length from the same point.
Could the result be that 30,60, 90 degree triangle? (if one connected the points?
| | | |______________________ (line one = 2X perpendicular)
This would seem to suggest that the formula for a 30, 60, 90 degree triangle would be that side B = 2A and the hypotenuse = SQRT ((B*B)+(A*A))
Could it be this simple?

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On Mon, 10 Apr 2006 15:07:38 GMT, "Gooey TARBALLS"

NO. As was pointed out the two arms [at right angles] of a 30,60 right triangle are in the ratio 1:sqrt(3). The Hypotenuse is the side twice the shorter.
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| | >Draw a vertical line desired length of shortest leg (x). | >Draw a horizontal line 2x in length from the same point. | > | >Could the result be that 30,60, 90 degree triangle? (if one connected the | >points? | | NO. As was pointed out the two arms [at right angles] of a 30,60 | right triangle are in the ratio 1:sqrt(3). The Hypotenuse is the side | twice the shorter. |
You sure about this??
Seems to me the ratio of the sides of a scalene right triangle (30,60,90) is
1:2:sqrt3.
and, according to Pythagorus,
The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other 2 sides.
Give all this balderdash, were one to lay out a horizontal base line of 2" and erect a vertical line at one end of the base line that is 1" in height, the distance between the two unattached line ends would be exactly sqrt 3" long. Additionally, the smallest angle would be exactly 30 degrees, the larger angle would be exactly 60 degrees and the biggest angle would be exactly 90 degrees.
-- PDQ
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PDQ wrote:

erect a vertical line at one end of the base line that is 1" in height, the distance between the two unattached line ends would be exactly sqrt 3" long. Additionally, the smallest angle would be exactly 30 degrees, the larger angle would be exactly 60 degrees and the biggest angle would be exactly 90 degrees.
So, you're saying 1^2 + 2^2 = 3 ?
er
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| > | | > | >Draw a vertical line desired length of shortest leg (x). | > | >Draw a horizontal line 2x in length from the same point. | > | > | > | >Could the result be that 30,60, 90 degree triangle? (if one connected the | > | >points? | > | | > | NO. As was pointed out the two arms [at right angles] of a 30,60 | > | right triangle are in the ratio 1:sqrt(3). The Hypotenuse is the side | > | twice the shorter. | > | | > | > You sure about this?? | > | > Seems to me the ratio of the sides of a scalene right triangle (30,60,90) is | > | > 1:2:sqrt3. | > | > and, according to Pythagorus, | > | > The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other 2 sides. | > | > Give all this balderdash, were one to lay out a horizontal base line of 2" and erect a vertical line at one end of the base line that is 1" in height, the distance between the two unattached line ends would be exactly sqrt 3" long. Additionally, the smallest angle would be exactly 30 degrees, the larger angle would be exactly 60 degrees and the biggest angle would be exactly 90 degrees. | | So, you're saying 1^2 + 2^2 = 3 ? | | er | -- | email not valid
There are days when there seems to be a dis-connection between my extremities.
Must be age related --- I meant sqrt 5 and my fingers weren't listening.
To get a side of sqrt 3 one would need the hypotenuse to be exactly 2" long.
The height of an iscoceles triangle of this side length would be sqrt 3. -- PDQ
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and....???
. . . sqrt(3) . . 2 . . . . ......... 1
Not well drawn, but the left side and base form the right angle.
2^2 = [sqrt(3)]^2 + 1^2
4 = 3 + 1
Yes, I'm sure about this. It is *not* the two that are at right angles that are one twice the other, and so far as I know, "horizontal" and "vertical" are still at right angles. If that was the case, you'd have 1^2 + 2^2 = 1+4 = 5, and the hypotenuse would be sqrt(5) ...and, more importantly here, the angles would NOT then be 30,60.
Figure also this way: The 30,60 triangle is 1/2 of an equilateral triangle. You drop a perpendicular to one side of the equilateral, dividing that side in two. Look at the drawing to see that it is the hypotenuse that is a side of the original equilateral triangle. If each side was 2, the divided base in 1.
OK, that's it. Any more and I'll have to charge you for my time.
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Nope.
Give all this balderdash, were one to lay out a horizontal base line of 2" and erect a vertical line at one end of the base line that is 1" in height, the distance between the two unattached line ends would be exactly sqrt 3" long. Additionally, the smallest angle would be exactly 30 degrees, the larger angle would be exactly 60 degrees and the biggest angle would be exactly 90 degrees.
--
PDQ

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NO. The line joining those ends is the hypotenuse. Never mind the other relationships, but consider that the hypotenuse is *always* the longest side of a right triangle [very simple to prove]. Drag out your calculator if you have to, but sqrt(3) is about 1.7321, which is still less than 2 where I come from.
In your diagram the length would be sqrt(1^2 + 2^2) = sqrt(5), and the angles would not be 30,60.
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Gooey TARBALLS wrote:

The lengths of each arm are 1 and sqrt(3), the hypotenuse is length 2. That was also given in austrobis' geometric proof.
er
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On Mon, 10 Apr 2006 05:30:35 GMT, Australopithecus scobis

etc..
If using straightedge and ruler, draw a horizontal line, and mark off two equal distances AB, BC. From A draw an arc [no need to change distances on the compasses] to cover half way [but above] from A to B. using B, draw the same arc back towards A. Suppose the arc meet at D. Draw a line through AD. Widen the compasses. Darw an arc from A above B. Draw the same arc from C above B to meet the first in E. Draw the line through BE.
Angle BAD will be 60 degrees. Angle ABE will be 90. There's your 30.60,90 triangle.
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[[.. sneck ..]]
and I was still doing it the hard way.
the _easy_ way:
1) Draw a straight line. 2) grab a compass, set it to any convenient value. 3) put the point on the line, and draw a 180-degree arc, where both ends are on the line. 4) put the compass point on one end of the arc from step 3 5) strike a new arc, intersecting the arc from step 3 6) connect that intersection point to the two ends of the 180-degree arc.
Voila! a 30-60-90 right triangle.
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