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03/06/2021 12:19 pm
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In the figure, ΔODC ∝ 1/4 ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
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03/06/2021 12:20 pm
As we can see from the figure, DOB is a straight line.
Therefore, ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° – 125° (Given, ∠BOC = 125°)
= 55°
In ΔDOC, sum of the measures of the angles of a triangle is 180º
Therefore, ∠DCO + ∠CDO + ∠DOC = 180°
⇒ ∠DCO + 70º + 55º = 180°(Given, ∠CDO = 70°)
⇒ ∠DCO = 55°
It is given that, ΔODC ∝ 1/4 ΔOBA,
Therefore, ΔODC ~ ΔOBA.
Hence, Corresponding angles are equal in similar triangles
∠OAB = ∠OCD
⇒ ∠OAB = 55°
∠OAB = ∠OCD
⇒ ∠OAB = 55°