Question

if two sides of a triangle are unequal prove that the angle opposite to the longer side is greater​

Answer 1

Answer:

If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater). You may prove this theorem by taking a point P on BC such that CA = CP. ... Now, take a point P on line AB such that AP = AC. Join the two points C and P to set CP.

Step-by-step explanation:

PLEASE FOLLOW ME

Answer 2

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

if two sides of a triangle are unequal prove that the angle opposite to the longer side is greater

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

${\blue{\sf\underline{Given}}}$

A ∆ABC in which AC > AB.

${\orange{\sf\underline{To\:prove}}}$

∠ABC > ∠BCA.

${\blue{\sf\underline{Construction}}}$

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

${\orange{\sf\underline{Proof}}}$

we know that in a triangle, the angles opposite to equal sides are equal.

So, in ∆ABD, we have

AB = AD = ∠BDA = ∠ABD.

Now,in ∆BCD,side CD hsd been produced to A, forming exterior angle ∠BDA.

∴ ∠BDA > ∠BCD⠀[exterior angle is greater than int.opp.angle]

⇒ ∠BDA = ∠BCA⠀[∴ ∠BCD = ∠BCA]

⇒ ∠ABD > ∠BCA⠀[using (i)]

⇒∠ABC > ∠ABD > ∠BCA⠀[∴ ∠ABC > ∠ABD]

⇒∠ABC > ∠BCA.

Hence, ∠ABC > ∠BCA.

━━━━━━━━━━━━━━━━━━━━━━━━━