Question 2:
In the given figure sides AB and AC of ∆ABC are extended to points P and Q
respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Answer:
In the 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 larger angle is larger.)
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