Question

12. Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle.
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Answer 1

Answer:

Consider △PQR where PR is the longest side

So we get PR>PQ

i.e. ∠Q>∠R..(1)

We also know that PR>QR

i.e. ∠Q>∠P..(2)

By adding both the equations

∠Q+∠Q>∠R+∠P

So we get

2∠Q>∠R+∠P

By adding ∠Q on both LHS and RHS

2∠Q+∠Q>∠R+∠P+∠Q

We know that ∠R+∠P+∠Q=180

∘

So we get

3∠Q>180

∘

By division

∠Q>60

∘

So we get

∠Q>

3

2

(90

∘

)

i.e. ∠Q>

3

2

of a right angle

Therefore, it is proved that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater

3

2

of a right angle.

solution

Answer 2

Answer:

let AB be the longest side then,

= AB > BC andAB > CA

=∠C >∠A and ∠C > ∠B

{therefor angle opposite to longer side is large}

= 2∠C >(∠A+∠B)

3∠C > (∠A+∠B+∠C)

= 3 ∠C > 180°