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  • Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 2 divided 3 of a right angle

Prove that in a triangle, other than an equilateral triangle, the angle opposite the longest side is greater than \frac{2}{3} that of a right angle.

Answers (1)

Given: ABC is a triangle and AC is the longest side

To Prove: \angle ABC > \frac{2}{3}\times 90^{\circ}

Proof: AC is the longest side

\Rightarrow\angle B > \angleC                                                    (angle opposite to longest side is great)

\Rightarrow\angleB > \angleA

By adding both

\angleB + \angleB > \angleC + \angleA

2\angleB > \angleC +\angleA

2\angleB + \angleB > \angleC + \angleA + \angleB             (adding \angleB to LHS and RHS)

3\angleB >\angleC + \angleA + \angleB

3\angleB > 180°                                          (Sum of all interior angles in a triangle is 180°)

\angleB > 60°

\Rightarrow \angle B > \frac{2}{3}\times 90^{\circ}

Hence proved.

Posted by

infoexpert24

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