Question

The side BC of ∆ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD = 115°, find ∠ALC.

Answer 1

$$\begin{gathered} \angle \mathrm{ACD}\mathrm{and}\angle \mathrm{ACL}\mathrm{make}\mathrm{a}\mathrm{linear}\mathrm{pair}. \\ \therefore \angle \mathrm{ACD}+\angle \mathrm{ACB}=180° \\ \Rightarrow 115°+\angle \mathrm{ACB}=180° \\ \angle \mathrm{ACB}=180°-115° \\ \angle \mathrm{ACB}=65° \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Therefore},\mathrm{for}△\mathrm{ABC},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180° \\ \Rightarrow 30°+\angle \mathrm{BAC}+65°=180° \\ Or, \\ \angle \mathrm{BAC}=85° \\ =\angle \mathrm{LAC}=\frac{\angle \mathrm{BAC}}{2}=\frac{85°}{2} \\ \mathrm{Using}\mathrm{the}\mathrm{above}\mathrm{rule}\mathrm{for}△\mathrm{ALC},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{ALC}+\angle \mathrm{LAC}+\angle \mathrm{ACL}=180° \\ \Rightarrow \angle \mathrm{ALC}+\frac{85°}{2}+65°=180°(\because \angle \mathrm{ACL}=\angle \mathrm{ACB}) \\ Or, \\ \angle \mathrm{ALC}=180°-\frac{85°}{2}-65° \\ =\mathbf{\angle }\mathbf{ALC}\mathbf{=}\frac{\mathbf{145}\mathbf{°}}{\mathbf{2}}\mathbf{=}\mathbf{72}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{°} \\ \mathrm{Thus}, \\ \angle ALC=72\frac{1}{2}° \end{gathered}$$