Question

In given figure, AM $\perp$ BC and AN is angle bisector of $\angle$BAC then the value of $\angle$MAN is

• A. ${7.5}^{\circ }$
• B. $5^{\circ }$
• C. ${10}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

Answer: (B)

$5^{\circ }$

Since, in $\Delta ABC$:

$\angle A+\angle B+\angle C=180°$

$⟹\angle A=180°-55°-65°=60°$.

Now, $AN$ is angle bisector of $\angle A$ then,

$\angle CAN=\angle BAN=30°$

Now, $\angle ANM=\angle NCA+\angle CAN=55°+30°=85°$........(Sum of two interior angles is equal to opposite exterior angle).

Now, in $\Delta ANM$:

$\angle AMN=90°$ and

$\angle MAN+\angle ANM+\angle AMN=180°$

$⟹\angle MAN=180°-90°-85°=5°$