ABCD – rhombus, CK perpendicular to the plane of the rhombus, CK = 2√3 cm, AB = 4 cm,

ABCD – rhombus, CK perpendicular to the plane of the rhombus, CK = 2√3 cm, AB = 4 cm, Angle BAD = 60 degrees. Find the angle between planes (ABK) and (ABCD)?

The angle between the planes ABK and ABCD will be the angle Between the perpendiculars CH and KN, that is, the KNS angle.
1. Find the area of ​​the rhombus ABCD by the formula:
S = a ^ 2 * sinA,
where a is the length of the side of the rhombus, A is the angle between two adjacent sides of the rhombus.
Since in a rhombus all sides are equal, then the area ABCD is equal to:
S = AB ^ 2 * sinBAD;
S = 4 ^ 2 * sin60 = 16 * √3 / 2 = 16√3 / 2 = 8√3 (cm ^ 2).
2. Also, the area of ​​a rhombus can be calculated through the side and the height drawn to it:
S = a * h.
Then:
h = S / a.
In the rhombus given by the condition, the CH height is equal to:
CH = 8√3 / 4 = 2√3 (cm).
3. Consider a triangle KCH: angle KCH = 90 degrees (since CK is the perpendicular to the plane ABCD), CK = 2√3 cm and CH = 2√3 cm – legs. Since the legs СK and СН are equal, then КСН is an isosceles right-angled triangle with sides СK and СН, the base is КН, and the angles СKН and КНС are angles at the base, therefore the angle СKН = angle КНС = х.
By the theorem on the sum of the angles of a triangle:
angle СKН + angle KНС + angle KСН = 180 degrees;
x + x + 90 = 180;
2x = 180 – 90;
2x = 90;
x = 90/2;
x = 45.
angle СKН = angle KНС = x = 45 degrees.
Answer: KНС angle = 45 degrees.