Question
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
Consider $\triangle\text{ABC}$ in which $BC$ is the longest side. To prove $\angle\text{A}=\frac{2}{3}$ right angle Proof In $\triangle\text{ABC},\ \text{BC}>\text{AB}.$
$[$consider $BC$ is the largest side$]$
$\Rightarrow\ \angle\text{A}>\angle\text{C} ...(i) [$angle opposite the lngest side is greatest$]$
$\text{and}\ \text{BC}>\text{AC}$
$\Rightarrow\ \angle\text{A}>\angle\text{B}, [$angle opposite the longest side is greatest$]$
On adding Eqs. $(i)$ and $(ii),$ we get $2\angle\text{A}>\angle\text{B}+\angle\text{C}$
$\Rightarrow\ 2\angle\text{A}+\angle\text{A}>\angle\text{A}+\angle\text{C}$ [adding $\angle\text{A}$ both sides]
$\Rightarrow\ 3\angle\text{A}>\angle\text{A}+\angle\text{B}+\angle\text{C}$
$\Rightarrow\ 3\angle\text{A}>180^\circ [$sum of the angles of a triangle is $180^\circ ]$
$\Rightarrow\ \angle\text{A}>\frac{2}{3}\times90^\circ$
i.e., $\angle\text{A}>\frac{2}{3}$ of a right angle Hence proved.
$[$consider $BC$ is the largest side$]$
$\Rightarrow\ \angle\text{A}>\angle\text{C} ...(i) [$angle opposite the lngest side is greatest$]$
$\text{and}\ \text{BC}>\text{AC}$
$\Rightarrow\ \angle\text{A}>\angle\text{B}, [$angle opposite the longest side is greatest$]$
On adding Eqs. $(i)$ and $(ii),$ we get $2\angle\text{A}>\angle\text{B}+\angle\text{C}$
$\Rightarrow\ 2\angle\text{A}+\angle\text{A}>\angle\text{A}+\angle\text{C}$ [adding $\angle\text{A}$ both sides]
$\Rightarrow\ 3\angle\text{A}>\angle\text{A}+\angle\text{B}+\angle\text{C}$
$\Rightarrow\ 3\angle\text{A}>180^\circ [$sum of the angles of a triangle is $180^\circ ]$
$\Rightarrow\ \angle\text{A}>\frac{2}{3}\times90^\circ$
i.e., $\angle\text{A}>\frac{2}{3}$ of a right angle Hence proved.
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