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Question 20
Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 23 of a right angle.

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Solution


ΔABC in which BC is the longest side.
In ΔABC,BC>AB,
[consider BC is the longest side]
A>C ...(i)
[angle opposite the longest side is greatest]
and BC > AC
A>B. ...(ii)
[angle opposite the longest side is greatest]
On adding Eqs. (i) and (ii), we get
2A>B+C
2A+A>A+B+C [adding A both sides]
3A>A+B+C
3A>180 [sum of angles of a triangle is 180]
A>23×90
i.e.,A>23of a right angle.


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