I need some, (ok a lot) of help in constructing a 30-60-90 triangle using classical construction methods, right now just about every thing anyone could offer will be a big help. I'm really quite lost,

Thanks in advance,
Son Of Dwight

You can do this easily by merely constructing the perpendicular bisector of a given line and connecting points.

Lay out a base and open your compass to a convenient size. From some point A on the base, mark off points B and C on the base equidistant from A. Open your compass to the length BC and strike intersecting arcs on either side of the base BC from B. Repeat from point C. Label the intersections of these pairs of arcs D above the base and E below, and connect them. The triangles BDC and BEC will be equilateral triangles, and the triangles BDA, BEA, ADC and AEC will all be triangles of 30, 60, and 90 degrees as required. I hope this is clear enough without a diagram.

[This message was edited by maiku on 10-21-02 at 03:39 PM.]

Draw line AB.
Draw arc AB.
Draw arc EAC center B.
Draw arc FBD center A.
Label intersection AC-BD G.
Label intersection AE-BF H.
Draw line GH.
Label intersection AB-GH I.
Draw arc BJ center I intersecting GH.
Draw arc IK center J.
Draw line KB.
Voila!

Thanks a million! I was finnaly able to get it done. My math book didn't do a very good job on exeplaning how to create the classical constructs.

Son of Dwight

Forget the arc AB. My mistake.