Construct a ∆ABC in which BC = 5.6 cm,∠B = 30° and the difference between the other two sides is 3 cm.

(i) First, draw the base, BC = 5.6 cm

(ii) Taking B as centre and some radius, draw an arc of a circle which intersects BC at point P.

(iii) Take P as centre and draw an arc of same radius in step (ii) which intersects the previous arc at point Q.

(iv) Now, take P and Q as centres respectively and draw arcs of a radius more than 1/2 PQ which intersect each other at R.

(v) Draw a ray BX passing through R. Thus,∠XBC = 30°.

(vi) Cut a line segment BM = 3 cm from BX.

(vii) Join MC.

(viii) Draw the right bisector of MC, which meet SX at A.

(ix) Join AC. Then, ∆ABC is the required triangle.