8.
Prove that in a triangle, other than an equilateral
triangle, angle opposite to the longest side is
2
greater than
of a right angle.
Answers
Answered by
2
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
Step-by-step explanation:
Answered by
1
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°
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