2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution: It is given that ∠PBC < ∠QCB We know that ∠ABC + ∠PBC = 180° So, ∠ABC = 180°-∠PBC Also, ∠ACB +∠QCB = 180° Therefore ∠ACB = 180° -∠QCB Now, since ∠PBC < ∠QCB, ∴ ∠ABC > ∠ACB Hence, AC > AB as sides opposite to the larger angle is always larger.
