Question

In the given figure, $AM\perp BC$ and $AN$ is the bisector of $\angle A$. Then $\angle MAN$ is

• A. $32{\frac{1}{2}}^0$
• B. $16{\frac{1}{2}}^0$
• C. ${16}^0$
• D. ${32}^0$

Answer 1

The correct option is C ${16}^0$

In $△ABM$,
$\angle ABM+\angle AMB+\angle MAB=180$
$65+90+\angle MAB=180$
$\angle MAB={25}^{\circ }$
In $△AMC$,
$\angle AMC+\angle ACM+\angle CAM=180$
$90+33+\angle CAM=180$
$\angle CAM={57}^{\circ }$
Let $\angle MAN=x$
we know, $AN$ bisects $\angle A$,
$\angle BAN=\angle NAC$
$\angle BAM+\angle MAN=\angle MAC-\angle MAN$
$25+x=57-x$
$2x=32$
$x={16}^{\circ }$