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

1.	In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 115° and ∠CDO = 70°. Find:(i) ∠DOC(ii) ∠DCO(iii) ∠OAB(iv) ∠OBA
Answer» 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)

In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 115° and ∠CDO = 70°. Find:(i) ∠DOC(ii) ∠DCO(iii) ∠OAB(iv) ∠OBA

Answer»

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)