Draw `angleBXA = 110^(@)` with the help of a protractor. Now , we use the following steps for required construction
(i) Taking X as centre and any radius daw an arc to intersect the rays XA and XB, say at E and D, respectively.
(ii) Taking D and E as centres and with the radius more than `(1)/(2)DE`, draw arcs to intersect each other say at F.
(iii) Draw the ray XF.
Thus , say XF is the required bisector of the angle BXA.
On measuring each angle , we get `angleBXC = angleAXC = 55^(@)`
`[becauseangleBXC = angleAXC = (1)/(2)angleBXA = (1)/(2)xx110^(@) = 55^(@)]`
