To construct a triangle ABC in which AB = 3.6cm, AC = 3.0cm and BC = 4.8cm, use the following steps.
(i) Draw a line segment BC of length 4.8cm
(ii) From B, point A is at a distance of 3.6cm. So, having B as the centre, draw an arc of radius 3.6cm.
(iii) From C, point A is at a distance of 3cm. So, having C as the centre, draw an arc of radius 3cm which intersect the previous arc at A.
(iv) Join AB and AC, thus,
ΔABC is the required triangle.
Here, angle B is smallest, as AC is the smallest side (Angle opposite to the smallest side is smaller)
To bisect angle B, we use the following steps.
(i) Taking B as the centre, we draw arcs intersecting AB and BC at D and E, respectively.
(ii) Taking D and E as centres we draw arcs intersecting at P.
(iii) Joining BP, we obtain angle bisector of
∠B (iv) Here,
∠ABC=39∘ Thus,
∠ABP=∠PBC=12×39∘=19.5∘