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
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Answer
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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