The acute angle of a rhombus whose side is a mean proportional between its diagonals is

Options

A. `15^(@)`
B. `20^(@)`
C. `30^(@)`
D. `80^(@)`

Explanation

To find the acute angle of a rhombus whose side is a mean proportional between its diagonals, we can follow these steps:

### Step 1: Understand the properties of the rhombus
In a rhombus, the diagonals bisect each other at right angles. Let's denote the diagonals as \( AC = x \) and \( BD = y \). The sides of the rhombus are equal, and we denote the side length as \( a \).

### Step 2: Set up the mean proportional relationship
Since the side \( a \) is the mean proportional between the diagonals, we have the relationship:
\[ a^2 = \frac{x \cdot y}{4} \]
This is derived from the fact that the area of the rhombus can also be expressed as:
\[ \text{Area} = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times x \times y \]
And since the area can also be expressed in terms of the side and the angle, we can relate these two expressions.

### Step 3: Use the Pythagorean theorem
In triangle \( DOC \) (where \( O \) is the intersection of the diagonals), we can apply the Pythagorean theorem:
\[ a^2 = \left(\frac{x}{2}\right)^2 + \left(\frac{y}{2}\right)^2 \]
This simplifies to:
\[ a^2 = \frac{x^2}{4} + \frac{y^2}{4} \]
Multiplying through by 4 gives:
\[ 4a^2 = x^2 + y^2 \]

### Step 4: Relate the two equations
From the mean proportional relationship, we have:
\[ 4a^2 = xy \]
Now we can set the two expressions for \( 4a^2 \) equal to each other:
\[ x^2 + y^2 = xy \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ x^2 - xy + y^2 = 0 \]
This is a quadratic equation in \( x \) or \( y \).

### Step 6: Solve for the ratio of the diagonals
Let \( t = \frac{y}{x} \), then \( y = tx \). Substituting into the equation gives:
\[ x^2 - x(tx) + (tx)^2 = 0 \]
Simplifying leads to:
\[ x^2(1 - t + t^2) = 0 \]
Thus, we have:
\[ 1 - t + t^2 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} \]
This indicates that \( t \) must be real, leading us to consider the values of \( t \) that yield acute angles.

### Step 8: Find the acute angle
Using the tangent function, we find:
\[ \tan \theta = \frac{y}{x} \]
From our earlier calculations, we can deduce that the acute angle \( \theta \) can be represented as:
\[ \tan \theta = 2 - \sqrt{3} \]
This corresponds to \( \theta = 15^\circ \).

### Final Result
Thus, the acute angle of the rhombus is:
\[ \boxed{30^\circ} \]