If `sin 3A = cos(A - 26^@)`, where 3A is an acute angle, find the value of A.

Answer

To solve the equation \( \sin 3A = \cos(A - 26^\circ) \), where \( 3A \) is an acute angle, we can follow these steps:

### Step 1: Use the co-function identity
We know that \( \sin x = \cos(90^\circ - x) \). Therefore, we can rewrite \( \sin 3A \) as:
\[ \sin 3A = \cos(90^\circ - 3A) \]
So, we can set up the equation:
\[ \cos(90^\circ - 3A) = \cos(A - 26^\circ) \]

### Step 2: Set the angles equal
Since the cosines are equal, we can set the angles equal to each other:
\[ 90^\circ - 3A = A - 26^\circ \]

### Step 3: Solve for A
Now, we will solve for \( A \):
1. Rearranging the equation gives:
\[ 90^\circ + 26^\circ = A + 3A \]
\[ 116^\circ = 4A \]

2. Dividing both sides by 4:
\[ A = \frac{116^\circ}{4} = 29^\circ \]

### Step 4: Conclusion
Thus, the value of \( A \) is:
\[ \boxed{29^\circ} \]