In a rhombus ABCD if `angleACB=40^(@),` then fine `angleADC.`

Answer

To solve the problem, we need to find the angle \( \angle ADC \) in rhombus \( ABCD \) given that \( \angle ACB = 40^\circ \).

### Step-by-Step Solution:

1. **Identify the properties of the rhombus**:
- In a rhombus, all sides are equal, and the opposite angles are equal.

2. **Recognize triangle \( ABC \)**:
- Since \( ABCD \) is a rhombus, triangles \( ABC \) and \( ACB \) are isosceles because \( AB = AC \) (sides of the rhombus).

3. **Set up the angles in triangle \( ABC \)**:
- We know that \( \angle ACB = 40^\circ \).
- Let \( \angle CAB = \angle ABC = x \) (since \( ABC \) is isosceles).

4. **Use the triangle angle sum property**:
- The sum of angles in triangle \( ABC \) is \( 180^\circ \):
\[ \angle CAB + \angle ABC + \angle ACB = 180^\circ \]
Substituting the known values:
\[ x + x + 40^\circ = 180^\circ \]
This simplifies to:
\[ 2x + 40^\circ = 180^\circ \]

5. **Solve for \( x \)**:
- Rearranging gives:
\[ 2x = 180^\circ - 40^\circ \]
\[ 2x = 140^\circ \]
\[ x = 70^\circ \]
- Therefore, \( \angle CAB = \angle ABC = 70^\circ \).

6. **Find \( \angle ADC \)**:
- Since opposite angles in a rhombus are equal, we have:
\[ \angle ADC = \angle ABC = 70^\circ \]

### Final Answer:
Thus, \( \angle ADC = 100^\circ \).