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Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsTriangles4 Marks
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.
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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°]

$\Rightarrow\ \angle\text{A}>\frac{2}{3}\times90^\circ$

i.e., $\angle\text{A}>\frac{2}{3}$ of a right angle Hence proved.

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