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In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB. - Mathematics

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In the given figure 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

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.)

Concept: Inequalities in a Triangle
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APPEARS IN

NCERT Class 9 Maths
Chapter 7 Triangles
Exercise 7.4 | Q 2 | Page 132

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