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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 $\angle MAN$.
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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, $\angle \mathrm{MAN}=16^{\circ}$.

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Updated on: 10-Oct-2022

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