The Total solution for NCERT class 6-12
Inthe given figure sides AB and AC of ΔABC are extended to points P and Qrespectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Inthe given figure,
∠ABC + ∠PBC = 180° (Linear pair)
⇒ ∠ABC = 180° − ∠PBC… (1)
Also,
∠ACB + ∠QCB = 180°
∠ACB = 180° − ∠QCB … (2)
As∠PBC < ∠QCB,
⇒ 180º − ∠PBC > 180º − ∠QCB
⇒ ∠ABC > ∠ACB[From equations (1) and (2)]
⇒ AC > AB (Side opposite to the largerangle is larger.)