Question

Question 20
Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle.

Answer 1

$\Delta ABC$ in which BC is the longest side.
In $\Delta ABC,BC>AB,$
[consider BC is the longest side]
$\Rightarrow \angle A>\angle C...\left(i\right)$
[angle opposite the longest side is greatest]
and BC > AC
$\Rightarrow \angle A>\angle B....\left(ii\right)$
[angle opposite the longest side is greatest]
On adding Eqs. (i) and (ii), we get
$2\angle A>\angle B+\angle C$
$\Rightarrow 2\angle A+\angle A>\angle A+\angle B+\angle C$ [adding $\angle A$ both sides]
$\Rightarrow 3\angle A>\angle A+\angle B+\angle C$
$\Rightarrow 3\angle A>{180}^{\circ }$ [sum of angles of a triangle is ${180}^{\circ }$]
$\Rightarrow \angle A>\frac{2}{3}\times {90}^{\circ }$
$i.e.,\angle A>\frac{2}{3}ofarightangle.$