If `tan(A+B)=sqrt(3)andtan(A-B)=(1)/(sqrt(3)),angleA+angleBlt90^(@)AgeB` , then `angleA` is

Options

A. `90^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ \tan(A + B) = \sqrt{3} \]
\[ \tan(A - B) = \frac{1}{\sqrt{3}} \]

2. **Identifying Angles**:
From trigonometric values, we know:
\[ \tan(60^\circ) = \sqrt{3} \quad \text{and} \quad \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Therefore, we can set:
\[ A + B = 60^\circ \quad \text{(1)} \]
\[ A - B = 30^\circ \quad \text{(2)} \]

3. **Solving the System of Equations**:
Now, we can add equations (1) and (2):
\[ (A + B) + (A - B) = 60^\circ + 30^\circ \]
This simplifies to:
\[ 2A = 90^\circ \]
Dividing both sides by 2 gives:
\[ A = 45^\circ \]

4. **Finding B**:
Now, substitute \( A = 45^\circ \) back into equation (1):
\[ 45^\circ + B = 60^\circ \]
Solving for \( B \):
\[ B = 60^\circ - 45^\circ = 15^\circ \]

5. **Conclusion**:
We have found:
\[ A = 45^\circ \quad \text{and} \quad B = 15^\circ \]
Since \( A + B < 90^\circ \) and \( A > B \), both conditions are satisfied.

Thus, the value of \( \angle A \) is:
\[ \boxed{45^\circ} \]