Question

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

 

Answer 1

It is given that DB is a straight line.
Therefore,
∠𝐷𝑂𝐶+ ∠𝐶𝑂𝐵=180°
∠𝐷𝑂𝐶=180°−115°=65°
(ii) In Δ DOC, we have:
∠𝑂𝐷𝐶+ ∠𝐷𝐶𝑂+ ∠𝐷𝑂𝐶=180°
Therefore,
700+ ∠𝐷𝐶𝑂+65°=180°
⟹ ∠𝐷𝐶𝑂=180−70−65=45°
(iii) It is given that Δ ODC - Δ OBA
Therefore,
∠𝑂𝐴𝐵= ∠𝑂𝐶𝐷=45°
(iv) Again, Δ ODC- Δ OBA
Therefore,
∠𝑂𝐵𝐴= ∠𝑂𝐷𝐶=70°