Question

Side BC of ∆ABC has been produced to D on left hand side and to E on right hand side such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?
(a) 65°
(b) 75°
(c) 50°
(d) 55°

Answer 1

(b) 75°

We have :
$$\begin{gathered} \mathbf{\therefore }\angle ABD+\angle ABC=180°\left[\because DE\text{is a straight line}\right] \\ \Rightarrow 125°+\angle ABC=180° \\ \Rightarrow \angle ABC=55° \end{gathered}$$
Also,
$$\begin{gathered} \angle ACE+\angle ACB=180° \\ \Rightarrow 130°+\angle ACB=180° \\ \Rightarrow \angle ACB=50° \\ \therefore \angle BAC+\angle ABC+\angle ACB=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow \angle BAC+55°+50°=180° \\ \Rightarrow \angle BAC=75° \\ \Rightarrow \angle A=75° \end{gathered}$$