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In the figure, $AM \perp BC$ and $AN$ is the bisector of $\angle A$. If $\angle B = 65^o$ and $\angle C = 33^o$, find $\angle MAN$.
Given:
$AM \perp BC$ and $AN$ is the bisector of $\angle A$.
$\angle B = 65^o$ and $\angle C = 33^o$.
To do:
We have to find $\angle MAN$.
Solution:
$\mathrm{AM} \perp \mathrm{BC}$
This implies,
$\angle \mathrm{AMC}=90^{\circ}$
$\angle \mathrm{AMN}=90^{\circ}$
In $\triangle \mathrm{ABC}$,
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\angle \mathrm{A}+65^{\circ}+33^{\circ}=180^{\circ}$
$\angle \mathrm{A}+98^{\circ}=180^{\circ}$
$\angle \mathrm{A}=180^{\circ}-98^{\circ}=82^{\circ}$
$\mathrm{AN}$ is the bisector of $\angle \mathrm{A}$.
This implies,
$\angle \mathrm{NAC}=\angle \mathrm{NAB}=\frac{1}{2} \angle \mathrm{A}$
$=\frac{1}{2} \times 82^{\circ}$
$=41^{\circ}$
In $\triangle \mathrm{AMN}$,
$\angle \mathrm{ANM}=\angle \mathrm{C}+\angle \mathrm{NAC}$
$=33^{\circ}+41^{\circ}$
$=74^{\circ} \)
In $\triangle \mathrm{MAN}$,
$\angle \mathrm{MAN}+\angle \mathrm{AMN}+\angle \mathrm{ANM}=180^{\circ}$
$\angle \mathrm{MAN}+90^{\circ}+74^{\circ}=180^{\circ}$
$\angle \mathrm{MAN}+164^{\circ}=180^{\circ}$
$\angle \mathrm{MAN}=180^{\circ}-164^{\circ}$
$\angle \mathrm{MAN}=16^{\circ}$.
Hence, $\angle \mathrm{MAN}=16^{\circ}$.