Question

In the given figure, sides CB and BA of ∆ABC have been produced to D and E, respectively, such that ∠ABD = 110° and ∠CAE = 135°. Then ∠ACB = ?
(a) 35°
(b) 45°
(c) 55°
(d) 65°

Answer 1

(d) 65°

We have :
$$\begin{gathered} \angle ABD+\angle ABC=180°\left[\because CBD\text{is a straight line}\right] \\ \Rightarrow 100°+\angle ABC=180° \\ \Rightarrow \angle ABC=70° \end{gathered}$$
Side AB of triangle ABC is produced to E.
$$\begin{gathered} \therefore \angle CAE=\angle ABC+\angle ACB \\ \Rightarrow 135°=70°+\angle ACB \\ \Rightarrow \angle ACB=65° \end{gathered}$$