If `tan (A+B) = sqrt3` and `tan(A-B)=1/sqrt3,0^(@)ltA+Ble90^(@),A gt B` , find A and B .

Answer

To solve the problem, we have the following equations based on the given information:

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: Identify angles from tangent values

From the first equation, we know:

\[ \tan(A + B) = \sqrt{3} \implies A + B = 60^\circ \quad \text{(since } \tan 60^\circ = \sqrt{3}\text{)} \]

### Step 2: Write the first equation

Thus, we have our first equation:

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

### Step 3: Analyze the second equation

From the second equation, we have:

\[ \tan(A - B) = \frac{1}{\sqrt{3}} \implies A - B = 30^\circ \quad \text{(since } \tan 30^\circ = \frac{1}{\sqrt{3}}\text{)} \]

### Step 4: Write the second equation

So, we have our second equation:

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

### Step 5: Solve the system of equations

Now, we can solve these two equations simultaneously. We will add Equation 1 and Equation 2:

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

This simplifies to:

\[ 2A = 90^\circ \]

Dividing both sides by 2 gives:

\[ A = 45^\circ \]

### Step 6: Substitute to find B

Now, we can substitute \( A = 45^\circ \) back into either equation to find \( B \). Using Equation 1:

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

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

\[ B = 15^\circ \]

### Step 7: Verify the conditions

Now we check the conditions:

1. \( A + B = 45^\circ + 15^\circ = 60^\circ \) (which is between \( 0^\circ \) and \( 90^\circ \)).
2. \( A > B \) (since \( 45^\circ > 15^\circ \)).

Both conditions are satisfied.

### Final Answer

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

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