Question

The sides BC, CA and AB of ∆ABC have been produced to D, E and F, respectively, as shown in the figure, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Then, ∠ACD + ∠BAE + ∠CBF = ?
(a) 240°
(b) 300°
(c) 320°
(d) 360°

Answer 1

(d) 360°

We have :
$$\begin{gathered} \angle 1+\angle BAE=180°...\left(i\right) \\ \angle 2+\angle CBF=180°...\left(ii\right)\text{and} \\ \angle 3+\angle ACD=180°...\left(iii\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Adding}\left(i\right),\left(ii\right)\mathrm{and}\left(iii\right),\text{we get:} \\ \left(\angle 1+\angle 2+\angle 3\right)+\left(\angle BAE+\angle CBF+\angle ACD\right)=540° \end{gathered}$$
$$\begin{gathered} \Rightarrow 180°+\angle BAE+\angle CBF+\angle ACD=540°\left[\because \angle 1+\angle 2+\angle 3=180°\right] \\ \Rightarrow \angle BAE+\angle CBF+\angle ACD=360° \end{gathered}$$