The height of the rhombus is 2√6, and the cosine of the angle between its height and the smaller

The height of the rhombus is 2√6, and the cosine of the angle between its height and the smaller diagonal is 0.6. Find the area of the rhombus.

Given: ABCD – rhombus;
AN = 2 * √6 – rhombus height;
cos (NAC) = 0.6.
Find: S =? Is the area of the rhombus.

Let us calculate what the first diagonal of the rhombus AC is equal to:
AC = h / cos (NAC) = 2 * √6 / 0.6 = √6 * 10/3 ≈ 8.165

Let’s calculate NC:
AC² = AN² + NC²;
NC² = 6 * 100/9 – 4 * 6 = 42.67;
NC = 6.53.

Let’s calculate the tangent of the angle NCA:
tg (NCA) = AN / NC = 2 * √6 / 6.53 = 0.75.

Find half of the second diagonal:
BO = 0.5 * AC * tg (NCA) = 0.5 * √6 * 10/3 * 0.75 ≈ 3.062.

Let’s define the area of a rhombus through its diagonals:
S = 8.165 * 3.062 = 25 sq. units