Geometry help needed

[PDQ]

| > | >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 | >=20 | > You sure about this?? | >=20 | > Seems to me the ratio of the sides of a scalene right triangle = (30,60,90) is | >=20 | > 1:2:sqrt3. | >=20 | > and, according to Pythagorus,=20 | >=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. | >=20 | > 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 | So, you're saying 1^2 + 2^2 =3D 3 ? |=20 | er | --=20 | 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.

--=20 PDQ

[Guess who]

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.

[Joe Gorman]

I must be lazy. After getting 2 points on the circle, which, combined with the center point, gives me a 60° angle and a side of 1 unit, I extended the other side of the angle to 2 units. Connecting this endpoint with the end of the 1 unit side, on the circle, gives me the side that's the square root of 3 units. Joe

[Robert Bonomi]

[[.. 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.

[Australopithecus scobis]

The point, of course, is that 30:60:90 triangles are trivial to generate with geometry. OP has his pick, so far, of at least three different methods.

I just use a plastic drafting triangle.