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Question

In the given figure, ΔODCΔOBA, BOC=115o and CDO=70o.

Find (i) DOC

(ii) DCO

(iii) OAB

(iv) OBA.


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Solution

GIVEN:
∆ODC ~ ∆OBA, ∠BOC =115° & ∠CDO = 70 °

DOB is a straight line and OC is a ray on it.
∠DOC + ∠ COB = 180°
∠DOC = 180° – 115° = 65°
∠DOC = 65°

In ΔDOC,
∠DCO + ∠ CDO + ∠ DOC = 180°
[Sum of the measures of the angles of a ∆ is 180º]
∠DCO + 70º + 65º = 180°
∠DCO + 135° = 180°
∠DCO = 180° - 135°
∠DCO = 45°

ΔODC ~ ΔOBA [given]
∠OAB = ∠OCD
[Corresponding angles are equal in similar triangles.]
∠ OAB = 45°

∠OBA = ∠ODC [Corresponding angles are equal in similar triangles.]
∠OBA = 70°

Hence, ∠DOC = 65° , ∠DCO = 45°, ∠ OAB = 45°, ∠OBA = 70°.



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