If `sin2A=cos(A-12^(@))and2A` is an acute angle , find the value of A.

Answer

To solve the equation \( \sin 2A = \cos(A - 12^\circ) \) given that \( 2A \) is an acute angle, we can follow these steps:

### Step 1: Use the co-function identity
We know that:
\[ \sin x = \cos(90^\circ - x) \]
Thus, we can rewrite \( \sin 2A \) as:
\[ \sin 2A = \cos(90^\circ - 2A) \]
Now, we can set our equation:
\[ \cos(90^\circ - 2A) = \cos(A - 12^\circ) \]

### Step 2: Set the angles equal
Since the cosines are equal, we can set the angles equal to each other:
\[ 90^\circ - 2A = A - 12^\circ \]

### Step 3: Rearrange the equation
Now, we will rearrange the equation to isolate \( A \):
\[ 90^\circ + 12^\circ = A + 2A \]
\[ 102^\circ = 3A \]

### Step 4: Solve for \( A \)
Now, divide both sides by 3:
\[ A = \frac{102^\circ}{3} = 34^\circ \]

### Conclusion
Thus, the value of \( A \) is:
\[ \boxed{34^\circ} \]