Question
In the given figure, $\text{AM}\perp\text{BC}$ and $AN$ is the bisector of $\angle\text{A}.$ If $\angle\text{ABC}=70^\circ$ and $\angle\text{ACB}=20^\circ,$ find $\angle\text{MAN}.$
✓
Answer
In $\triangle\text{ABC},$ by angle sum property,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+70^\circ+20^\circ=180^\circ$
$\Rightarrow\angle\text{A}=90^\circ$
In $\triangle\text{ABM},$ by angle sum property,
$\angle\text{BAM}+\angle\text{ABM}+\angle\text{AMB}=180^\circ$
$\Rightarrow\angle\text{BAM}+70^\circ+90^\circ=180^\circ$
$\Rightarrow\angle\text{BAM}=20^\circ$
Since $AN$ is the bisector of $\angle\text{A},$
$\angle\text{BAN}=\frac{1}{2}\angle\text{A}$
$\Rightarrow\angle\text{BAN}=\frac{1}{2}\times90^\circ=45^\circ$
Now, $\angle\text{MAN}+\angle\text{BAM}=\angle\text{BAN}$
$\Rightarrow\angle\text{MAN}+20^\circ=45^\circ$
$\Rightarrow\angle\text{MAN}=25^\circ$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+70^\circ+20^\circ=180^\circ$
$\Rightarrow\angle\text{A}=90^\circ$
In $\triangle\text{ABM},$ by angle sum property,
$\angle\text{BAM}+\angle\text{ABM}+\angle\text{AMB}=180^\circ$
$\Rightarrow\angle\text{BAM}+70^\circ+90^\circ=180^\circ$
$\Rightarrow\angle\text{BAM}=20^\circ$
Since $AN$ is the bisector of $\angle\text{A},$
$\angle\text{BAN}=\frac{1}{2}\angle\text{A}$
$\Rightarrow\angle\text{BAN}=\frac{1}{2}\times90^\circ=45^\circ$
Now, $\angle\text{MAN}+\angle\text{BAM}=\angle\text{BAN}$
$\Rightarrow\angle\text{MAN}+20^\circ=45^\circ$
$\Rightarrow\angle\text{MAN}=25^\circ$
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