Question

In the given figure, $\Delta ODC\sim \Delta OBA,\angle BOC={115}^{o}and\angle CDO={70}^o$.

Find (i) $\angle DOC$

(ii) $\angle DCO$

(iii) $\angle OAB$

(iv) $\angle OBA$.

Answer 1

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°.