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°