In a triangle BCD, BC = CD = a cm, and one of the acute angles is 60, then the triangle BCD is equilateral.
The height НМ of the rhombus ABCD is equal to the height DH of the equilateral triangle BCD, since НМ and DH are perpendicular to BC and AD.
DH is the height and median of the triangle BCD, then CH = BH = BC / 2 = a / 2 cm.
In a right-angled triangle BDH tg60 = DH / BH.
DН = ВН * tg60 = a * √3 / 2 cm.
Answer: The height of the rhombus is a * √3 / 2 cm.
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