Abcd is a rhombus such that ∟adb = 50°, then what is the measure of ∟acb?
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ABCD is a Rhombus.
∠ADB = 50°
As AD and BC are parallel, and BD is a transversal line,
∠DBC = ∠ADB = 50°
Diagonals AC and BD intersect at right angles at O. So in ΔOCD,
∠ACB = ∠OCB = 90° - 50° = 40°
∠ADB = 50°
As AD and BC are parallel, and BD is a transversal line,
∠DBC = ∠ADB = 50°
Diagonals AC and BD intersect at right angles at O. So in ΔOCD,
∠ACB = ∠OCB = 90° - 50° = 40°
kvnmurty:
:-)
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