Construct a triangle ABC, having given AB = 4.8 cm, AC = 4 cm and ∠A = 75°. Find a point P

(i) inside the triangle ABC

(ii) outside the triangle ABC

equidistant from B and C; and at a distance of 1.2 cm from BC.

Steps of construction :

1. Draw a line segment AB = 4.8 cm.

2. At A, draw a ray AX making an angle of 75°.

3. Cut off AC = 4 cm from AX.

4. Join BC. ABC is the required triangle.

5. Draw two lines l and m parallel to BC at a distance of 1.2 cm

6. Draw the perpendicular bisector of BC which intersect l and m at P and P'.

(i) Hence, P is the point inside the triangle ABC which is equidistant from B and C; and at a distance of 1.2 cm from BC.

(ii) Hence, P' is the point outside the triangle ABC which is equidistant from B and C; and at a distance of 1.2 cm from BC.