ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB = _________.

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Solution

Let ∠ADB be x.

As we know, angles formed by the diagonals of a rhombus is 90°

In ∆BOC,
∠OBC + ∠BCO + ∠COB = 180°
⇒ ∠OBC + 40° + 90° = 180°
⇒ ∠OBC + 130° = 180°
⇒ ∠OBC = 180° − 130°
⇒ ∠OBC = 50°

Since, AD || BC
Thus, ∠ADB = ∠DBC (alternate interior angles)
⇒ x = 50°

Hence, ∠ADB = 50°.

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