Question

prove that if two sides of triangle is unequal the angle opposite to the longer side is larger.
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Answer 1

The solution of this question is in the attachment...

Answer 2

Step-by-step explanation:

Given: A ΔABC in which AC>AB (say)

To prove: ∠ABC>∠ACB

Construction: Mark a point D on AC such that AB=AD. Join BD.

Proof: In ΔABD

AB=AD (by construction)

∠1=∠2 …(i) (angles opposite to equal sides are equal)

Now in ΔBCD

∠2>∠DCB (ext. angle is greater than one of the opposite interior angles)

∠2>∠ACB …(ii) [∵∠ACB=∠DCB]

From (i) and (ii), we get

∠1>∠ACB …(iii)

But ∠1 is a part of ∠ABC

∠ABC>∠1 …..(iv)

Now from (iii) and (iv), we get

∠ABC>∠ACB

Hence proved.