The measure of each of two opposite angles of a rhombus is `60^(@)` and the measure of one of its sides is 10 cm. The length of its smaller diagonal is :

Options

A. 10 cm
B. `10sqrt(3) cm`
C. `10sqrt(2) cm`
D. `(5)/(2)sqrt(2) cm`

Answer

To find the length of the smaller diagonal of the rhombus, we can follow these steps:

### Step 1: Understand the properties of the rhombus
A rhombus has four equal sides and opposite angles that are equal. The diagonals of a rhombus bisect each other at right angles.

### Step 2: Identify the given information
- Each of two opposite angles of the rhombus is \(60^\circ\).
- The length of each side of the rhombus is \(10 \, \text{cm}\).

### Step 3: Determine the angles of the rhombus
Since opposite angles are equal, the other two angles will be:
\[ 180^\circ - 60^\circ = 120^\circ \]
So, the angles of the rhombus are \(60^\circ, 120^\circ, 60^\circ, 120^\circ\).

### Step 4: Use the properties of triangles
When we draw the diagonals of the rhombus, they intersect at right angles. Let's denote the vertices of the rhombus as \(A, B, C, D\) such that \( \angle A = 60^\circ \) and \( \angle B = 120^\circ \).

### Step 5: Split the rhombus into triangles
By drawing the diagonals \(AC\) and \(BD\), we can form two triangles \(ABD\) and \(BCD\). We focus on triangle \(ABD\).

### Step 6: Apply the sine rule in triangle \(ABD\)
In triangle \(ABD\):
- Side \(AB = 10 \, \text{cm}\)
- Angle \(ABD = 60^\circ\)
- Angle \(ADB = 120^\circ\)

We can use the sine rule to find the length of diagonal \(BD\):
\[ \frac{AB}{\sin(\angle ADB)} = \frac{BD}{\sin(\angle ABD)} \]
Substituting the known values:
\[ \frac{10}{\sin(120^\circ)} = \frac{BD}{\sin(60^\circ)} \]

### Step 7: Calculate the sine values
Using the known sine values:
\[ \sin(120^\circ) = \frac{\sqrt{3}}{2}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Substituting these into the equation:
\[ \frac{10}{\frac{\sqrt{3}}{2}} = \frac{BD}{\frac{\sqrt{3}}{2}} \]
This simplifies to:
\[ \frac{20}{\sqrt{3}} = \frac{BD}{\frac{\sqrt{3}}{2}} \]

### Step 8: Solve for \(BD\)
Cross-multiplying gives:
\[ BD \cdot \frac{\sqrt{3}}{2} = 20 \]
Thus,
\[ BD = \frac{20 \cdot 2}{\sqrt{3}} = \frac{40}{\sqrt{3}} \approx 10\sqrt{3} \, \text{cm} \]

### Conclusion
The length of the smaller diagonal \(BD\) is \(10\sqrt{3} \, \text{cm}\).