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Question

If one angle of a triangle is less then the sum of the other two, show that the triangle is acute angled.

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Solution

Consider ABC as a triangle
According to the question it can be written as
A<B+C
Add A to both the sides of the question
So we get
A+A<A+B+C
We know that the sum of all the angles in a triangle is 180.
So we can write it as
A+B+C=180
So we get
2<A<180
By division we get
A<180/2
A<90
In the same way we can also write
B<A+C
Add B to both the sides of the equation
So we get
B+B<A+B+C
We know that the sum of all the angles in a triangle is 180.
So we can write it as
A+B+C=180
So we get
2B<180
By division we get
B<180/2
B<90
So we know that
C<A+B
Add C to both the sides of the equation
So we get
C+C<A+B+C
We know that the sum of all the angles in a triangle is 180.
So we can write it as
A+B+C=180
So we get
2C<180
By division we get
C<180/2
C<90
Therefore, it is proved that the triangle ABC is acute angled.

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