Suppose that you have a line L and some point A on L:

How will you construct a ray (or line) through A which is inclined at 60^{0} to L?

**Step 1:** Taking A as center and any radius, draw an arc which intersects L at B**:**

**Step 2:** Now, taking B as center and AB as radius, draw another arc which intersects the first arc at C:

**Step 3:** Draw a ray (or line) through A and C. This will be inclined at 60^{0} to L**:**

Why does this construction work?

**Proof:** Note that AB = AC, since these are radii of the same circular arc. Also, BC = BA, since these too are radii of the same (second) circular arc. Thus,

AB = BC = AC

This means that \(\Delta ABC\) is equilateral, and so, \(\angle BAC\) = 60^{0}.

Note that you can construct an angle of 30^{0} by bisecting an angle of 60^{0}, and you can further construct an angle of 15^{0} by bisecting an angle of 30^{0}.