Question

In Fig., ABC is a right triangle right angled at A. D lies on BA produced and DE ⊥ BC, intersecting AC at F. If ∠AFE = 130°, find

(i) ∠BDE
(ii) ∠BCA
(iii) ∠ABC

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Here}, \\ \angle \mathrm{BAF}+\angle \mathrm{FAD}=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow \angle \mathrm{FAD}=180°-\angle \mathrm{BAF}=180°-90°=90° \\ \mathrm{Also}, \\ \angle \mathrm{AFE}=\angle \mathrm{ADF}+\angle \mathrm{FAD}\left(\text{Exterior angle property}\right) \\ \angle \mathrm{ADF}+90°=130° \\ \angle \mathrm{ADF}=130°-90°=40° \\ \left(\mathrm{ii}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Therefore},\mathrm{for}∆\mathrm{BDE},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{BDE}+\angle \mathrm{BED}+\angle \mathrm{DBE}=180°. \\ \Rightarrow \angle \mathrm{DBE}=180°-\angle \mathrm{BDE}-\angle \mathrm{BED}=180°-90°-40°=50°...\left(\mathrm{i}\right) \\ \mathrm{Also}, \\ \angle \mathrm{FAD}=\angle \mathrm{ABC}+\angle \mathrm{ACB}\left(\text{Exterior angle property}\right) \\ \Rightarrow 90°=50°+\angle \mathrm{ACB} \\ \mathrm{Or}, \\ \angle \mathrm{ACB}=90°-50°=40° \\ \left(\mathrm{iii}\right)\angle \mathrm{ABC}=\angle \mathrm{DBE}=50°[\mathrm{From}(\mathrm{i}\left)\right] \end{gathered}$$