If `tan2A=cot(A-18^(@))` where 2A is an acute angle, find the value of A.

Answer

To solve the equation \( \tan 2A = \cot (A - 18^\circ) \), we can follow these steps:

### Step 1: Rewrite the cotangent function
We know that \( \cot \theta = \tan (90^\circ - \theta) \). Therefore, we can rewrite the equation as:
\[ \tan 2A = \tan (90^\circ - (A - 18^\circ)) \]

### Step 2: Simplify the right-hand side
Now simplify the right-hand side:
\[ \tan 2A = \tan (90^\circ - A + 18^\circ) = \tan (108^\circ - A) \]

### Step 3: Set the angles equal
Since \( \tan \theta_1 = \tan \theta_2 \) implies \( \theta_1 = \theta_2 + n \cdot 180^\circ \) (where \( n \) is an integer), we can set:
\[ 2A = 108^\circ - A + n \cdot 180^\circ \]
However, since \( 2A \) is an acute angle, we will consider \( n = 0 \):
\[ 2A = 108^\circ - A \]

### Step 4: Solve for A
Now, we can solve for \( A \):
\[ 2A + A = 108^\circ \]
\[ 3A = 108^\circ \]
\[ A = \frac{108^\circ}{3} = 36^\circ \]

### Conclusion
Thus, the value of \( A \) is:
\[ \boxed{36^\circ} \]