Question

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

(i) ∠BDE

(ii) ∠BCA

(iii) ∠ABC

Answer 1

(i) Here,

∠BAF + ∠FAD = 180° (Linear pair)

∠FAD = 180° – ∠BAF = 180° – 90° = 90°

Also from the figure,

∠AFE = ∠ADF + ∠FAD (Exterior angle property)

∠ADF + 90° = 130°

∠ADF = 130° − 90° = 40°

(ii) We know that the sum of all the angles of a triangle is 180°.

Therefore, for △BDE, we have

∠BDE + ∠BED + ∠DBE = 180°

∠DBE = 180° – ∠BDE

∠BED = 180° – 90° – 40° = 50° …. Equation (i)

Again from the figure we have,

∠FAD = ∠ABC + ∠ACB (Exterior angle property)

90° = 50° + ∠ACB

∠ACB = 90° – 50° = 40°

(iii) From equation we have

∠ABC = ∠DBE = 50°