Draw an angle of `110^(@)` with the help of a protractor and bisect it. Measure each of these angles. Justify your answer.

Explanation

STEPS OF CONSTRUCTION
(i) Draw `angleAOB=110^(@)` with the help of a protractor.
(ii) With `O` as centre and a convenient radius, draw an arc cutting `OA` at `P` and `OB` at `Q.`
(iii) With `P` as centre and a convenient radius, draw an arc.
(iv) With `Q` as centre and with the same radius, draw another arc, cutting the previous arc at a point `C.`
(v) Join `OC` and produce it beyond `C`.
Then, `OC` is the required bisector of `angleAOB`

Verification :
On measuring `angleAOC` and `angleBOC` you will find that
`angleAOC=55^(@)` and `angleBOC=55^(@)`.
Thus, `angleAOC=angleBOC=55^(@)` and therefore, OC is the bisector of `angleAOB`.
Justification :
Join CP and CQ.
In `Delta OPC` and `Delta OQC`, we have
`OP=OQ` (radii of the same arc)
`PC=QC` (arcs of equal radii)
`OC=OC` (common)
`:. Delta OPC cong Delta OQC`.
Hence, `anglePOC=angleQOC` (c.p.c.t.).
Consequently, OC is the bisector of `angleAOB`.