Question

In the given figure, ∠BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Then ∠AED = ?
(a) 120°
(b) 100°
(c) 80°
(d) 110°

Answer 1

(a) 120°
From $∆ABC$, we have:
$$\begin{gathered} \angle BAC+\angle ABC+\angle ACB=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 30°+50°+\angle ACB=180° \\ \Rightarrow \angle ACB=100° \end{gathered}$$
Now,
$$\begin{gathered} \angle ACB+\angle ECD=180°\left[\because BCD\text{is a straight line}\right] \\ \Rightarrow 100°+\angle ECD=180° \\ \Rightarrow \angle ECD=80° \end{gathered}$$
Side CE of triangle CED is produced to A.
$$\begin{gathered} \therefore \angle AED=\angle ECD+\angle EDC[\mathrm{Exterior}\mathrm{angle}property] \\ \Rightarrow \angle AED=\left(80+40\right)° \\ \Rightarrow \angle AED=120° \end{gathered}$$