Question

In the given figure, BE ⊥ AC, ∠DAC = 30° and ∠DBE = 40°. Find ∠ACB and ∠ADB.

Answer 1

From $∆BEC$, we have:
$$\begin{gathered} \angle CBE+\angle BEC+\angle ECB=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 40°+90°+\angle ECB=180° \\ \Rightarrow \angle ECB=50° \\ \mathrm{Now},\angle ACB=\angle ECB=50° \end{gathered}$$
Side CD of triangle ACD is produced to B.
$$\begin{gathered} \therefore \angle ADB=\angle DAC+\angle ACD \\ \Rightarrow \angle ADB=30°+50°\left[\because \angle ACD=\angle ECB\right] \\ \Rightarrow \angle ADB=80° \end{gathered}$$