To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 cm. So, having B as centre, draw an arc of radius 3.6 cm. 3.From C, point A is at a distance of 3 cm. So, having C as centre, draw an arc of radius 3 cm which intersect previous arc at A. 4.Join AB and AC. Thus, ΔABC is the required triangle. Flere, angle B is smallest, as AC is the smallest side. To direct angle B, we use the following steps. 1.Taking B as centre, we draw an are intersecting AB and BC at D and E, respectively. 2.Taking D and E as centres we draw arcs intersecting at P. 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°