In the given figure, ΔODC ~ ΔOBA, ∠BOC = 115° and ∠CDO = 70°.

Find (i) ∠DOC (ii) ∠DCO (iii) ∠OAB (iv) ∠OBA.

(i) To find ∠DOC, we can observe the straight line DB.

∠DOC + ∠COB = 180° [∵ sum of all angles in a straight line is 180°]

⇒ ∠DOC + 115° = 180°

⇒ ∠DOC = 180° - 115°

⇒ ∠DOC = 65°

(ii) In ∆DOC,

And given that, ∠CDO = 70°, ∠DOC = 65° (from (i))

∠DOC + ∠DCO + ∠CDO = 180°

⇒ 65° + ∠DCO + 70° = 180°

⇒ ∠DCO + 135° = 180°

⇒ ∠DCO = 180° - 135°

⇒ ∠DCO = 45°

(iii) We have derived ∠DCO from (ii), ∠DCO = 45°

Thus, ∠OAB = 45° [∵ ∠OAB = ∠DCO as ∆ODC ∼ ∆OBA]

(iv) It’s given that, ∠CDO = 70°

Thus, ∠OBA = 70° [∵ ∠OBA = ∠CDO as ∆ODC ∼ ∆OBA]