![](https://www.sarthaks.com/?qa=blob&qa_blobid=11625017851579095829)
Construction:
Steps (1) Draw a line segment BC = 5.6 cm.
Steps (2) At B, draw BE such that ∠CBE = 60°.
Steps (3) At B draw BF such that ∠EBF = 90°.
Steps (4) Draw ⊥r bisector to BC, which intersects
BF at 0.
Steps (5) With O as centre and OB as radius draw a circle.
Steps (6) From C, mark an arc of 4 cm on BC at D.
Steps (7) The ⊥r bisector intersects the circle at I.
Join ID.
Steps (8) ID produced meets the circle at A.
Now join AB and AC. ΔABC is the required triangle