In figure (3) given below, ABCD is a rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3 : 2, find the angles of the △AOD.

Let ∠OAB = 3x and ∠OBA = 2x. The diagonals of rhombus are perpendicular to each other. ∴ ∠AOB = 90° In △AOB, ⇒ ∠AOB + ∠OAB + ∠OBA = 180° ⇒ 90° + 3x + 2x = 180° ⇒ 5x = 90° ⇒ x = ⇒ x = 18°. ∠OAB = 3x = 3(18°) = 54°. ∠OBA = 2x = 2(18°) = 36°. Since, diagonals of rhombus bisect vertex angles. ∴ ∠OAD = ∠OAB = 54°, ∠AOD = 90° (The diagonals of rhombus are perpendicular to each other.) In △AOD, ⇒ ∠AOD + ∠OAD + ∠ODA = 180° ⇒ 90° + 54° + ∠ODA = 180° ⇒ ∠ODA + 144° = 180° ⇒ ∠ODA = 180° - 144° ⇒ ∠ODA = 36°. Hence, ∠ODA = 36°, ∠OAD = 54°, ∠AOD = 90°.

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Mathematics

In figure (3) given below, ABCD is a rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3 : 2, find the angles of the △AOD.

Rectilinear Figures

ICSE

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