Question
In Fig. $\text{AM}\perp\text{BC}$ and AN is the bisector of $\angle\text{A}.$ If $\angle\text{B}=65^\circ$ and $\angle\text{C}=33^\circ,$ find $\angle\text{MAN}.$
✓
Answer
Let $\angle\text{BAN}=\angle\text{NAC}=\text{x} $ $[\therefore\text{AN}\text{ bisects }\angle\text{A}]$$\therefore\angle\text{ANM}=\text{x}+33^\circ$ [Exterior angle property]
In $\triangle\text{AMB}$$\angle\text{BAM}=90^\circ-65^\circ=25^\circ$ [Exterior angle property]
$\therefore\angle\text{MAN}=\angle\text{BAN}-\angle\text{BAM}=(\text{x}-25)^\circ$
Now in $\triangle\text{MAN},$$(\text{x}-25)^\circ+(\text{x}+33)^\circ+90^\circ=180^\circ$ [Angle sum property]
$\Rightarrow2\text{x}+8^\circ=90^\circ$
$\Rightarrow2\text{x}=82^\circ$
$\Rightarrow\text{x}=41^\circ$
$\therefore\text{MAN}=\text{x}-25^\circ$
$=41^\circ-25^\circ$
$=16^\circ$
In $\triangle\text{AMB}$$\angle\text{BAM}=90^\circ-65^\circ=25^\circ$ [Exterior angle property]
$\therefore\angle\text{MAN}=\angle\text{BAN}-\angle\text{BAM}=(\text{x}-25)^\circ$
Now in $\triangle\text{MAN},$$(\text{x}-25)^\circ+(\text{x}+33)^\circ+90^\circ=180^\circ$ [Angle sum property]
$\Rightarrow2\text{x}+8^\circ=90^\circ$
$\Rightarrow2\text{x}=82^\circ$
$\Rightarrow\text{x}=41^\circ$
$\therefore\text{MAN}=\text{x}-25^\circ$
$=41^\circ-25^\circ$
$=16^\circ$
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