Question

In the given figure, $\angle$ABC = 65° and $\angle$ACB = 33°. If AM $\perp$ BC and AN is the bisector of $\angle$A, then $\angle$MAN = __.

• 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

${16}^{\circ }$

$\underset{–}{In\Delta ABC}$

$\angle BAC+{65}^{\circ }+{33}^{\circ }={180}^{\circ }...$(Sum of angle of a $\Delta$)

$\angle BAC=180-{98}^{\circ }$

= ${82}^{\circ }$

(ii) $\therefore \angle BAN=\angle NAC=\frac{{82}^{\circ }}{2}={41}^{\circ }$

(iii) $\underset{–}{In\Delta ANC}$

$\angle ANM={41}^{\circ }+{33}^{\circ }...$ (exteriar angle theorem)

= ${74}^{\circ }$

(iv) In $\Delta$ AMN

$\angle MAN+{90}^{\circ }+{74}^{\circ }={180}^{\circ }$ (Sum of Ls of a $\Delta )$

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

$={16}^{\circ }$

Aid to memory $\angle MAN=\frac{1}{2}({65}^{\circ }-{33}^{\circ })$