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 = $\dfrac{90°}{5}$

⇒ 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°.