If A and B are positive acute angles such that sin `(A - B) =1/2 `and `cos (A + B) = 1/2`, then A and B are given by

Options

A. `A=45^@ ,B=15^@`
B. ` A = 15^@ , B = 45^@`
C. ` A = 30^@ , B = 30^@ `
D. None of these

Answer

To solve the problem, we need to find the values of angles A and B given the equations:

1. \( \sin(A - B) = \frac{1}{2} \)
2. \( \cos(A + B) = \frac{1}{2} \)

### Step 1: Solve for \( A - B \)

From the first equation, we know that:

\[ \sin(A - B) = \frac{1}{2} \]

The sine function equals \( \frac{1}{2} \) at angles \( 30^\circ \) and \( 150^\circ \). However, since A and B are acute angles, we only consider:

\[ A - B = 30^\circ \]

### Step 2: Solve for \( A + B \)

From the second equation, we have:

\[ \cos(A + B) = \frac{1}{2} \]

The cosine function equals \( \frac{1}{2} \) at angles \( 60^\circ \) and \( 300^\circ \). Again, since A and B are acute angles, we only consider:

\[ A + B = 60^\circ \]

### Step 3: Set up the system of equations

Now we have a system of two equations:

1. \( A - B = 30^\circ \) (Equation 1)
2. \( A + B = 60^\circ \) (Equation 2)

### Step 4: Solve the system of equations

To solve for A and B, we can add both equations:

\[ (A - B) + (A + B) = 30^\circ + 60^\circ \]

This simplifies to:

\[ 2A = 90^\circ \]

Dividing both sides by 2 gives:

\[ A = 45^\circ \]

Now, substitute \( A = 45^\circ \) back into Equation 2 to find B:

\[ 45^\circ + B = 60^\circ \]

Subtracting \( 45^\circ \) from both sides gives:

\[ B = 15^\circ \]

### Step 5: Conclusion

Thus, the values of A and B are:

\[ A = 45^\circ, \quad B = 15^\circ \]

### Final Answer

So, the angles A and B are \( 45^\circ \) and \( 15^\circ \) respectively.

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