### Theory:

Important!
Construct an angle of $$120^\circ$$ at the initial point of a given ray and justify the construction.
Step 1: Draw a ray $$BC$$ with $$B$$ as an initial point.

Step 2: Taking $$B$$ as the centre and any convenient radius, draw an arc to intersect the ray $$BC$$ at $$D$$.

Step 3: With $$D$$ as the centre and same radius, draw an arc that cuts the previously drawn arc at $$E$$. Again, taking $$E$$ as the centre and the same radius, draw another arc which also cuts the previously drawn arc at $$F$$.

Step 4: Now, draw the ray $$BA$$ passing through $$F$$. Thus, $$\angle ABC = 120^\circ$$ is the required angle.

JUSTIFICATION:

Join $$BE$$, $$DE$$ and $$EF$$.

In triangle $$BED$$:

$$BE = BD = ED$$ [Radius of equal arcs by construction]

Thus, $$\Delta BED$$ is an equilateral triangle.

$$\Rightarrow \angle BED = \angle EDB = \angle DBE = 60^\circ$$

Now, in triangle $$BEF$$:

$$BE = BF = FE$$ [Radius of equal arcs by construction]

Thus, $$\Delta BEF$$ is an equilateral triangle.

$$\Rightarrow \angle BEF = \angle BFE = \angle EBF = 60^\circ$$

$$\angle DBF = \angle DBE + \angle EBF$$ (by construction)

$$\angle DBF = 60^\circ + 60^\circ$$

$$\angle DBF = 120^\circ$$

Therefore, $$\angle ABC = 120^\circ$$.