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In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB. Q.2

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The best way to solve the question of triangles form exercise 6.3 of class 10th math, how i solve in easy way In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

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  1. 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 ∝ ¼ ΔOBA,

    Therefore, ΔODC ~ ΔOBA.

    Hence, Corresponding angles are equal in similar triangles

    ∠OAB = ∠OCD

    ⇒ ∠ OAB = 55°

    ∠OAB = ∠OCD

    ⇒ ∠OAB = 55°

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