Question

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, find ∠MAN.
Figure

Answer 1

$$\begin{gathered} \mathrm{In}∆ABC,\mathrm{using}\mathrm{angle}-\mathrm{sum}\mathrm{property},\mathrm{we}\mathrm{have}: \\ \angle ABC+\angle BCA+\angle CAB=180° \\ \Rightarrow 70°+20°+\angle CAB=180° \\ \Rightarrow 90°+\angle CAB=180° \\ \Rightarrow \angle CAB=180°-90° \\ \Rightarrow \angle CAB=90° \\ \mathrm{Now},AN\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle A. \\ \therefore \angle BAN=\frac{1}{2}\angle CAB \\ =\frac{1}{2}\times 90° \\ =45° \\ \mathrm{In}∆ABM,\mathrm{using}\mathrm{angle}-\mathrm{sum}\mathrm{property},\mathrm{we}\mathrm{have}: \\ \angle ABM+\angle BMA+\angle MAB=180° \\ \Rightarrow 70°+90°+\angle MAB=180° \\ \Rightarrow 160°+\angle MAB=180° \\ \Rightarrow \angle MAB=180°-160°=20° \\ \mathrm{From}\text{the}\mathrm{figure}, \\ \angle MAN=\angle BAN-\angle MAB \\ =45°-20° \\ =25° \\ \mathrm{Therefore},\angle MAN=25° \end{gathered}$$