Question

in fig. 7.48 sides ab and ac of Δabc are extended to points p and q respectively .Also angle ∠pbc ab .

Answer 1

$\bf \huge {\underline {\underline \red{QuEsTiOn}}}$

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In fig. 7.48 sides AB and AC of ΔABC are extended to points P and Q respectively .Also angle ∠PBC < ∠QCB. Show that AC > AB.

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$\bf \huge {\underline {\underline \red{AnSwEr}}}$

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Given

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• ∠PBC < ∠QCB

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To Find

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• AC > AB

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Solution

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∠PBC + ∠ABC = 180° [ Linear Pair ]

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∠ABC = 180° - ∠PBC ────── i

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∠QCB + ∠ACB = 180° [ Linear Pair ]

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∠ACB = 180° - ∠QCB ────── ii

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∠PBC < ∠QCB [ given ]

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So, 180° - ∠PBC > 180° - ∠QCB

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From equation i and ii

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∠ABC > ∠ACB

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So, AC > AB [ Sides opposite to equal angles of a triangle are equal ]