abcd is a rhombus such that angle acb=40 find angle adb
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Answer:
50°
Step-by-step explanation:
ABCD is a rhombus. We know that diagonals of rhombus bisect each other perpendicularly. Let the point of bisect be O.
Hence, ∠BOC= 90°
∠OCB = 40° (Given)
AD∥BC and BD is the transversal --- (Opposite sides of rhombus are parallel to each other)
∴ ∠ADB = ∠DBC ---- (Alternate angles)
In △OBC,
∠BOC + ∠OCB + ∠OBC = 180°
⇒ 90° + 40° + ∠OBC = 180°
⇒ ∠OBC =180° - 130°
∴ ∠OBC = 50°
But ∠OBC =∠DBC
∴ ∠ADB = 50° ---( Alternate angle)
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