If `tan(A+B)=sqrt3` and `tan(A-B)=(1)/(sqrt3),0^(@)lt A+Ble90^(@), A gt B`, then find the value of A and B.

Answer

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

1. \( \tan(A + B) = \sqrt{3} \)
2. \( \tan(A - B) = \frac{1}{\sqrt{3}} \)

We also know that \( 0^\circ < A + B < 90^\circ \) and \( A > B \).

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

From the first equation, we have:

\[ \tan(A + B) = \sqrt{3} \]

The angle whose tangent is \( \sqrt{3} \) is \( 60^\circ \). Therefore, we can write:

\[ A + B = 60^\circ \quad \text{(Equation 1)} \]

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

From the second equation, we have:

\[ \tan(A - B) = \frac{1}{\sqrt{3}} \]

The angle whose tangent is \( \frac{1}{\sqrt{3}} \) is \( 30^\circ \). Therefore, we can write:

\[ A - B = 30^\circ \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

Now we have a system of two equations:

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

We can solve these equations by adding them together:

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

This simplifies to:

\[ 2A = 90^\circ \]

Dividing both sides by 2 gives:

\[ A = 45^\circ \]

### Step 4: Substitute to find \( B \)

Now, we can substitute the value of \( A \) back into Equation 1 to find \( B \):

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

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

\[ B = 15^\circ \]

### Conclusion

Thus, the values of \( A \) and \( B \) are:

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