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 30-60-90 triangles.
For your situation, the shortest side is 24"/1.732 = 13.857
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
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.
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.
Thanks for all the help. This is to make a rough triangle to check on building a one-time project. Not a high accuracy carpentry job this time.
Why not use geometry? For a 30° angle you can lay it out more easily by geometry.
"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.
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
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?
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.
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.| >Could the result be that 30,60, 90 degree triangle? (if one connected = the=20 | >points? |=20 | 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. |=20
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,=20
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.=20--=20 PDQ
the sum of the squares of the other 2 sides.
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
The lengths of each arm are 1 and sqrt(3), the hypotenuse is length 2. That was also given in austrobis' geometric proof.
er
the sum of the squares of the other 2 sides.
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 + 1Yes, 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.
In ASCII ? Not really !
You can do it either way. For a general solution to any angle, then trig would be the way to go - especially with cheap calculators. However
90°. 60° and 30° are easy constructions with just dividers and they're worth knowing.Draw a circle. With your dividers set on the same radius, step out this distance repeatedly around the circumference of the circle. It fits exactly 6 times, meaning that the radii to each of these points make angles of 60°.
To get from 60° to 30°, then you can either bisect it or you can subtract 60° from 90° (you'll often already have a right angle laid out for some other purpose).
To bisect the angle, use the dividers as before and mark out a point at approximately two radii from the centre, at the intersection of arcs drawn from two of the circumferential points you marked earlier. A radius drawn out to this point bisects the circumferential points, at an angle of 30°
Nope.
"PDQ" wrote in message news:Bxx_f.2418$ snipped-for-privacy@news20.bellglobal.com... 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.
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