prove that in a triangle, other than equilateral triangle,opposite angle to the longest side is greater than 2/3rd of a right angle
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An equilateral triangle is that whose each angle measures 60 degree.
So, 2/3 of a right angle is same as 60 degree.
According to the law of sines,
The angle opposite to the longest side will be greater angle.
In a triangle,
as we know that , the sum of three angles of a triangle = 180 degree.
where at least one of the angles must be equal to or at least should be 60 degree.
Also, the greater angle can't be 60 degree. It must be more than 60 degree.
So, 2/3 of a right angle is same as 60 degree.
According to the law of sines,
The angle opposite to the longest side will be greater angle.
In a triangle,
as we know that , the sum of three angles of a triangle = 180 degree.
where at least one of the angles must be equal to or at least should be 60 degree.
Also, the greater angle can't be 60 degree. It must be more than 60 degree.
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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°
Step-by-step explanation:
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