Question

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

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

Answer 1

The correct option is C

${16}^{\circ }$

In given figure,
$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }\text{(Angles of same triangle)}\Rightarrow \angle BAC+{65}^{\circ }+{33}^{\circ }={180}^{\circ }\Rightarrow \angle BAC={82}^{\circ }$

$\angle CAN=\frac{\angle BAC}{2}={41}^{\circ }\text{(Given that AN is an angle bisector)}.\text{Now,}\angle CAN+\angle ANC+\angle ACN={180}^{\circ }\text{(Angles of same triangle)}\Rightarrow {41}^{\circ }+\angle ANC+{33}^{\circ }={180}^{\circ }\Rightarrow \angle ANC={106}^{\circ }$

$\angle ANC+\angle ANM={180}^{\circ }\text{(Linear pair)}\Rightarrow {106}^{\circ }+\angle ANM={180}^{\circ }\Rightarrow \angle ANM={74}^{\circ }$

$\angle ANM+\angle AMN+\angle MAN={180}^{\circ }\Rightarrow {74}^{\circ }+{90}^{\circ }+\angle MAN={180}^{\circ }\Rightarrow \angle MAN={16}^{\circ }.$