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 identity for sine
We know that \( \sin \theta = \cos(90^\circ - \theta) \). Therefore, we can rewrite \( \sin 2A \):
\[ \sin 2A = \cos(90^\circ - 2A) \]
So, we can set the equation as:
\[ \cos(90^\circ - 2A) = \cos(A - 12^\circ) \]

### Step 2: Set the angles equal
Since the cosine function is equal for angles that differ by \( 360^\circ \) or are supplementary, we can set the angles equal:
\[ 90^\circ - 2A = A - 12^\circ \]

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

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

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