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.
Answer
Given,
∠PBC < ∠QCB
Now,
∠ABC + ∠PBC = 180°
⇒ ∠ABC = 180° - ∠PBC
also,
∠ACB + ∠QCB = 180°
⇒ ∠ACB = 180° - ∠QCB
Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Thus, AC > AB as sides opposite to the larger angle is larger.
