Theory:

Important!
To know how to construct an angle, click here.
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.
 
Pic 11.png
 
Step 2: Taking \(B\) as the centre and any convenient radius, draw an arc to intersect the ray \(BC\) at \(D\).
 
Pic 12.png
 
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 \(F\).
 
Pic 13.png
 
Step 4: Now, draw the ray \(BA\) passing through \(F\). Thus, \(\angle ABC = 120^\circ\) is the required angle.
 
Pic 14.png
 
JUSTIFICATION:
 
Join \(BE\), \(DE\) and \(EF\).
 
Pic 15.png
 
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, \(\angle ABC = 120^\circ\).