Question

In the given figure, ∠BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Find ∠AED.
Figure

Answer 1

Side BC of $∆$ABC is produced to D.
$$\begin{gathered} \therefore \angle ACD=\angle CAB+\angle ABC[\text{E}\mathrm{xterior}\mathrm{angle}property] \\ \Rightarrow \angle ACD=\left(30+50\right)° \\ \Rightarrow \angle ACD=80° \end{gathered}$$

Side CE of $∆$CED is produced to A.
$$\begin{gathered} \therefore \angle AED=\angle ECD+\angle EDC[\mathrm{Exterior}\mathrm{angle}property] \\ \Rightarrow \angle AED=\left(80+40\right)° \\ \Rightarrow \angle AED=120° \end{gathered}$$