In the figure, ΔODC ∝ \(\dfrac{1}{4}\) ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

In the figure, ΔODC ∝ 1/4 ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO  and ∠OAB. – Triangles – Forum

Asked by Pragya Singh | 1 year ago |  74

1 Answer

Solution :-

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 180°

Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180°

⇒ ∠DCO + 7 + 55°​​​​​​​ = 180°(Given, ∠ CDO = 70°)

⇒ ∠DCO = 55°

It is given that, ΔODC ∝ \( \dfrac{1}{4}\) ΔOBA,

Therefore, ΔODC ~ ΔOBA.

Hence, Corresponding angles are equal in similar triangles

∠OAB = ∠OCD

⇒ ∠ OAB = 55°

∠OAB = ∠OCD

⇒ ∠OAB = 55°

Answered by Abhisek | 1 year ago

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