Question

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I am here with a new question !!
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Prove the THEORAM

IF TWO SIDES OF A TRIANGLE ARE UNEQUAL , THE ANGLE OPPOSITE TO THE LONGER SIDE IS LARGER ( OR GREATER )

⭕ FULL EXPLANATION ⭕

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Answer 1

hey friend ! I would have given u every thing in the explanation but i have to go so i would just tell what to do see take an example to prove it lets consider a triangle of sides 3 cm, 4 cm and 5 cm

now they are not equal but if u would be in 11th u would know that sin 37°=3/5 and cos 37°=4/5 thus the angles would be 53°and 90° then see the larger side that 5 cm is with the angle of 90° thus with the help of trigonometry u would get it

hope it helps and u know trigonometry

if u get it how td do can u mark me the brainliest answer

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.

hope it will helpful!!!!