Question

prove that in a triangle other than equilateral triangle , angle opposite to longest side is greater then 2/3 of right angle

Answer 1

Consider a triangle that is almost equilateral. One angle must therefore exceed 60o and it is well known that the larger the angle, the larger is the side opposite, and the converse is also true.

Therefore the angle opposite the longest side must exceed 60o (which is, of course, therefore > two thirds of a 90o angle. Of course, actual rt triangles or obtuse triangles corroborate this assertion, also.

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°

Step-by-step explanation:

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