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Triangle And Its Angles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsTriangle And Its Angles5 Marks
Question
In a $\triangle\text{ABC},\angle\text{ABC}=\angle\text{ACB}$ and the bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ intersect at O such that $\angle\text{BOC}=120^\circ.$ Show that $\angle\text{A}=\angle\text{B}=\angle\text{C}=60^\circ.$
✓

Answer

Given,

In $\triangle\text{ABC},$
$\angle\text{ABC}=\angle\text{ACB}$ Dividing both sides by '2' $\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ACB}$
$\Rightarrow\angle\text{OBC}=\angle\text{OCB}$
$[\therefore\text{OB},\text{OC }\text{bisects }\angle\text{B}\text{ and }\angle\text{C}]$ Now, $\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}$
$\Rightarrow120^\circ-90^\circ=\frac{1}{2}\angle\text{A}$
$\Rightarrow30^\circ\times(2)=\angle\text{A}$
$\Rightarrow\angle\text{A}=60^\circ$ Now in $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ (Sum of all angle of a triangle) $60^\circ+2\angle\text{ABC}=180^\circ$
$[\therefore\angle\text{ABC}=\angle\text{ACB}]$
$\Rightarrow2\angle\text{ABC}=180^\circ-60^\circ$
$\Rightarrow\angle\text{ABC}=\frac{120^\circ}{2}=60^\circ$
$\Rightarrow\angle\text{ABC}=\angle\text{ACB}$
$\therefore\angle\text{ACB}=60^\circ$ Hence Proved.

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