If `0ltAltB ltpi, sin A+sinB=sqrt((3)/(2)) and cos A+cosB=(1)/(sqrt2),` then A =

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `22(1^(@))/(2)`

Answer

To solve the problem, we need to find the value of angle \( A \) given the equations involving sine and cosine of angles \( A \) and \( B \).

### Step-by-Step Solution:

1. **Given Equations**:
We have:
\[ \sin A + \sin B = \sqrt{\frac{3}{2}} \quad \text{(1)} \]
\[ \cos A + \cos B = \frac{1}{\sqrt{2}} \quad \text{(2)} \]

2. **Using Trigonometric Identities**:
We can express the sums of sine and cosine using the identities:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

3. **Substituting into Equations**:
Substitute the identities into equations (1) and (2):
\[ 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = \sqrt{\frac{3}{2}} \quad \text{(3)} \]
\[ 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{2}} \quad \text{(4)} \]

4. **Dividing Equations**:
Divide equation (3) by equation (4):
\[ \frac{\sin\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A+B}{2}\right)} = \frac{\sqrt{\frac{3}{2}}}{\frac{1}{2}} \implies \tan\left(\frac{A+B}{2}\right) = \sqrt{3} \]

5. **Finding \( A+B \)**:
Since \( \tan\left(\frac{A+B}{2}\right) = \sqrt{3} \), we know:
\[ \frac{A+B}{2} = \frac{\pi}{3} \implies A+B = \frac{2\pi}{3} \quad \text{(5)} \]

6. **Using Equation (4)**:
From equation (4):
\[ 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{2}} \]
Substitute \( \frac{A+B}{2} = \frac{\pi}{3} \):
\[ 2 \cos\left(\frac{\pi}{3}\right) \cos\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{2}} \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ 2 \cdot \frac{1}{2} \cos\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{2}} \implies \cos\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{2}} \]

7. **Finding \( A-B \)**:
From \( \cos\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{2}} \):
\[ \frac{A-B}{2} = \frac{\pi}{4} \implies A-B = \frac{\pi}{2} \quad \text{(6)} \]

8. **Solving for \( A \) and \( B \)**:
Now we have two equations:
\[ A + B = \frac{2\pi}{3} \quad \text{(5)} \]
\[ A - B = \frac{\pi}{2} \quad \text{(6)} \]
Adding (5) and (6):
\[ 2A = \frac{2\pi}{3} + \frac{\pi}{2} \]
To add, convert to a common denominator:
\[ 2A = \frac{4\pi}{6} + \frac{3\pi}{6} = \frac{7\pi}{6} \implies A = \frac{7\pi}{12} \]

### Final Answer:
Thus, the value of \( A \) is:
\[ A = \frac{7\pi}{12} \]