Question 15 Marks

In $\triangle\text{DEF},  \angle\text{D} = 60^\circ , \angle\text{E} = 70^\circ $ and the bisectors of $\angle\text{E}$ and $\angle\text{F}$ meet at $O$. Find $(i) \ \angle\text{F}  \ (ii) \ \angle\text{EOF}.$

Answer

$i.$ As we know,
$\angle\text{D}+\angle\text{E}+\angle\text{F}=180^{\circ} [$angle sum property of triangle$]$
$\Rightarrow60^{\circ}+70^{\circ}+\angle\text{F}=180^{\circ}$
$[\because\angle\text{D}=60^{\circ}  $ and $\angle\text{E}=70^{\circ}]$
$\Rightarrow\angle\text{F}=180^{\circ}-130^{\circ}$
$\Rightarrow \ \angle\text{F}=50^{\circ}$
$ii.$ Now, as $FO$ is the bisector of $\angle\text{F}.$
So, $\angle\text{EFO}=\frac{\angle\text{F}}{2}=\frac{50^{\circ}}{2}=25^{\circ}$
and $\angle\text{OEF}=\frac{\angle\text{E}}{2}=\frac{70^{\circ}}{2}=35^{\circ}$
$[\because\angle\text{D}=60^{\circ} $ and $\angle\text{E}=70^{\circ}]$
$ \triangle\text{EOF},$
$\angle\text{EOF}+\angle\text{OEF}+\angle\text{OFE}=180^{\circ} [$angle sum property of a triangle$]$
$\Rightarrow \angle\text{EOF}+35^{\circ}+25^{\circ}=180^{\circ}$
$\Rightarrow \angle\text{EOF}=180^{\circ}=60^{\circ}$
$\Rightarrow  \angle\text{EOF}=120^{\circ}$

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