Question

If each angle of a triangle is less than the sum of the other two show that the triangle is acute angled??

Answer 1

hey
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acute angles - angles whose measure is less than 90 degree

Given :
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each angle of a ∆
is less than the sum of the other two angles

<A < <B + <C ____(1)

<B < <A + <C____(2)

<C < <A + <B _____(3)

to proof : <A < 90 degree
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<B < 90 degree

<C < 90 degree

Proof:
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< A + <B + < C = 180 degree ( angle sum property of a ∆)

<B + <C = 180 - <A

from equation (1)
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<A < <B +< C
<A < 180 - <A.
2<A <180

<A <90

similarly
<A + <C = 180 - <B

from equation (2)
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<B < <A + <C
<B < 180 -<B
2<B < 180

<B < 90

again
<A +<B = 180 -<C

from equation(3)
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<C < <A + <B

<C < 180 -<C
2<C <180

<C < 90

hence

<A < 90

<B < 90

<C < 90

prooved
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Answer 2

Let ∠A, ∠B and ∠C be the interior angles of ΔABC.

It is given that each angle is less than the sum of the other two angles.

Consider, ∠A < ∠B + ∠C

⇒ ∠A < 180° – ∠A  [∠A + ∠B + ∠C = 180°]

⇒ 2∠A < 180°

⇒ ∠A < 90°

Thus, ∠A is an acute angle.

Infact, all the angles are acute.

Hence, ΔABC is an obtuse angled triangle.