Question

In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 115° and ∠CDO = 70°. Find:

(i) ∠DOC

(ii) ∠DCO

(iii) ∠OAB

(iv) ∠OBA

Answer 1

Here ∆ODC ~ ∆OBA, so

∠D = ∠B = 70°

∠C = ∠A

∠COD = ∠AOB

(i) But ∠DOC + ∠BOC = 180° (Linear pair)

∠DOC + 115°= 180°

∠DOC = 180° – 115° = 65°

(ii) ∠DOC + ∠CDO + ∠DCO = 180° (Angles of a triangle)

65° + 70° + ∠DCO = 180°

135° + ∠DCO = 180°

∠DCO = 180° – 135° = 45°

(iii) ∠AOB = ∠DOC = 65° (vertically opposite angles)

∠OAB = ∠DCO = 45° (Since ∆ODC ~ ∆OBA)

(iv) ∠OBA = ∠CDO = 70° (Since ∆ODC ~ ∆OBA)