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Question
In the given figure, ΔODC~ΔOBA, ∠BOC = 115° and ∠CDO = 700.
Find (i) ∠DCO (ii) ∠DCO (iii) ∠OAB (iv) ∠OBA.
Solution
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°
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