Constructions
58. The 60° Blueprint · Building perfect angles with compass & ruler
The constructed angle ∠COA always remains exactly 60°.
A 60° angle can be constructed exactly with compass and straightedge: from vertex O, draw an arc of radius r that crosses the base ray at B. With the same radius, an arc centred at B crosses the first arc at C. Then ∠COA is exactly 60°, because OB = BC = OC makes triangle OBC equilateral.
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For subscribers · Selina ICSE: Constructions
What this lesson covers
Try to break it
Try dragging O or A around. The whole construction scales with you, but ∠COA never budges from 60°. Why? Because we are secretly building an equilateral triangle!
How you build it
Construct a 60 degree angle.
- Mark point O — the vertex of the angle.
- Mark point A to the right of O.
- Draw the ray OA — the base of the angle.
- With centre O, draw an arc of any radius that crosses the base ray OA.
- Mark point B where the arc crosses the base ray OA.
- With the same radius and centre B, draw an arc that crosses the first arc.
- Mark point C where the two arcs cross.
- Draw the ray from O through C — the second arm of the angle. ∠COA is your perfect 60° angle.
The proof, step by step
Prove that the constructed angle ∠COA equals 60°.
- OB = OC = r (radii of the same arc centered at O)
- BC = r (radius of the arc centered at B)
- Therefore, OB = OC = BC
- Triangle OBC is equilateral
- ∠BOC = 60°
- Since B lies on OA, ∠COA = ∠BOC = 60°
Worked example
In the standard construction of a 60° angle using a compass and straightedge, why is it essential to draw arcs of the *same* radius from both O and B?
Equal radii guarantee OB = OC = BC, making triangle OBC equilateral. All angles in an equilateral triangle are 60°, so ∠BOC = 60°, which gives ∠COA = 60°.
- To make the arcs intersect at two distinct points
- To ensure triangle OBC is equilateral — correct
- To draw a perpendicular bisector of OA
- To create a 90° angle at O