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.
The short side is long side/2
The medium side is short side x square root of 3.
These only apply for 306090 triangles.
For your situation, the shortest side is 24"/1.732 = 13.857
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.
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.
"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.
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
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.
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?
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.

 >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
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
 > 
 >  >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 disconnection 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
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.
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.
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.
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.
[[.. 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 180degree 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 180degree arc.
Voila! a 306090 right triangle.
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