Question

In the figure, AM $\perp$ BC and AN is the bisector of $\angle A$. If $\angle B$=${65}^{\circ }$ and $\angle C={33}^{\circ }$, find $\angle MAN.$

Answer 1

In $\Delta ABC,\angle B={65}^{\circ },\angle C={33}^{\circ }$

AM$\perp$ BC and AN is bisector of $\angle A$

$\because AM\perp BC$

$\therefore \angle AMC={90}^{\circ }$

$\Rightarrow \angle AMN={90}^{\circ }$

In $\Delta ABC,$

$\angle A+\angle B+\angle C={180}^{\circ }$

(Sum fo angles of a triangle)

$\Rightarrow \angle A+{65}^{\circ }+{33}^{\circ }={180}^{\circ }$

$\Rightarrow \angle A+{98}^{\circ }={180}^{\circ }$

$\Rightarrow \angle A={180}^{\circ }-{98}^{\circ }={82}^{\circ }$

$\because$ AN the bisector of $\angle A$

$\therefore \angle NAC=\angle NAB=\frac{1}{2}\angle A=\frac{1}{2}\times {82}^{\circ }={41}^{\circ }$

In$\Delta ANC,$

Ext. $\angle ANM=\angle C+\angle NAC$

=${33}^{\circ }+{41}^{\circ }={74}^{\circ }$

In $\Delta MAN,$

$\angle MAN+\angle AMN+\angle ANM={180}^{\circ }$

(Angles of a triangle)

$\Rightarrow \angle MAN+{90}^{\circ }+{74}^{\circ }={180}^{\circ }$

$\Rightarrow \angle MAN+{164}^{\circ }={180}^{\circ }$

$\Rightarrow \angle MAN={180}^{\circ }-{164}^{\circ }={16}^{\circ }$

Hence, $\angle MAN={16}^{\circ }$