Question

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. Find the measure of ∠MAN.

Answer 1

From $∆ABC$, we have:
$$\begin{gathered} \angle A+\angle B+\angle C=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow \angle A+65°+30°=180° \\ \Rightarrow \angle A=85° \end{gathered}$$
Since AN is the bisector of $\angle A$, we have:
$$\begin{gathered} \angle NAB=\frac{1}{2}\angle A \\ \Rightarrow \angle NAB=\frac{1}{2}\left(85\right)° \\ \Rightarrow \angle NAB=42.5° \end{gathered}$$

From the right $∆AMB$, we have:
$$\begin{gathered} \angle B+\angle MAB=90° \\ \Rightarrow \angle MAB=90°-\angle B \\ \Rightarrow \angle MAB=\left(90-65\right)° \\ \Rightarrow \angle MAB=25° \\ \therefore \angle MAN=\angle NAB-\angle MAB \\ \Rightarrow \angle MAN=\left(42.5-25\right)° \\ \Rightarrow \angle MAN=17.5° \end{gathered}$$