- Construction of Angles
- How To Construct A 90 Degree Angle
- How To Construct A 60 Degree Angle
- Steps Of Construction Of A 60 Degree Angle Using a Compass
- How To Construct A 30 Degree Angle
- How To Construct A 120 Degree Angle Using A Compass

One of the most beautiful aspects of geometry is how neatly everything fits together. Knowing how to bisect a line segment means we know how to make a $90\xb0$ angle. Knowing an equilateral triangle has three $60\xb0$ angles makes an easy job of constructing a $60\xb0$ angle with only a compass, a pencil, and a straightedge. Everything fits together. You need nearly the same tools Euclid used, thousands of years ago:

- A drawing compass
- A straightedge
- A pencil
- Some paper

- On the paper, draw a line segment longer than the ray of a the $90\xb0$ angle you need.
- Locate two points on the line segment at either end.
- Open the drawing compass to extend a bit beyond half the distance of the line segment (do this visually; no need for numbers).
- Swing an arc above and below the line segment.
- Without changing the compass, relocate it to the other endpoint.
- Swing another arc above and below the line segment. The two arcs should cross above and below the line segment.

Connecting the arc intersection points with the straightedge will produce a neat, sure $90\xb0$ angle, with your transversal and the original line meeting at four right angles!

Something practically mystical surrounds $60\xb0$. It forms interior angles of equilateral triangles. It is exactly 1/6th of a circle. Half of it plus itself forms a right angle. You could go on and on.

Making a $60\xb0$ angle begins by remembering that equilateral triangle.

- Construct a line segment with a straightedge. Label its endpoints. In our drawing, we will call them $PointsO$ and $G$.
*[insert drawing of left-to-right line segment OG]* - Place the drawing compass needle on $PointO$ and adjust it to meet $PointG$.
- Swing an arc upward from $PointG$ high above the line segment.
- Without adjusting the compass, relocate the needle to $PointG$.
- Swing an arc upwards, so it crosses the first arc.
- Use the straightedge to construct a line segment from $PointO$ up to the point of intersection of the two arcs.
- Label that line segment's new endpoint $PointD$.

The angle created from $PointsD$ to $O$ to $G$, by striking three congruent lengths, is $60\xb0$. If you wanted, you could connect $PointsD$ and $G$ and form the equilateral triangle. Hey, $DOG$, you did it!

A $30\xb0$ angle is half of a $60\xb0$ angle. So, to draw a $30\xb0$, construct a $60\xb0$ angle and then bisect it.

First, follow the steps above to construct your $60\xb0$ angle.

Bisect the $60\xb0$ angle with your drawing compass, like this:

- Without changing the compass, relocate the needle arm to one of the points on the rays. Swing an arc on the inside of the angle.
- Without changing the compass, relocate the needle arm to the other ray's point. Swing an arc on the inside of the angle.
- Use the straightedge to connect the intersection of the two arcs with the vertex of the $60\xb0$ angle. That line segment is an angle bisector, yielding two $30\xb0$ angles.

Remember how we said everything in geometry fits together? What is the supplementary angle to an angle of $120\xb0$? In other words, what angle must we add to $120\xb0$ to get $180\xb0$?

Did you say $60\xb0$? Sure, a $120\xb0$ angle is the adjacent angle to any of the $60\xb0$ angles you already constructed!

To construct your $120\xb0$ angle, construct a $60\xb0$ angle and then extend one of its sides far past the vertex, like this:

*[insert animation of 60° angle constructed, then run out the side and highlight the 120° angle adjoining it]*

That angle beyond the $60\xb0$ angle is your $120\xb0$ angle.

None of your constructions required numbers or measurement. Euclid would be proud of you!

Instructor: **Malcolm M.**

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

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