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.

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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

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