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.

Given:
1. \( \sin A + \sin B = \sqrt{\frac{3}{2}} \)
2. \( \cos A + \cos B = \frac{1}{\sqrt{2}} \)

We also know that \( 0 < A, B < \pi \).

### Step 1: Use the identities for sine and cosine sums

We can use the identities for the sum of sine and cosine:
- \( \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \)
- \( \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \)

Let \( S = \frac{A + B}{2} \) and \( D = \frac{A - B}{2} \).

Then we can rewrite the equations as:
1. \( 2 \sin(S) \cos(D) = \sqrt{\frac{3}{2}} \)
2. \( 2 \cos(S) \cos(D) = \frac{1}{\sqrt{2}} \)

### Step 2: Solve for \( \cos(D) \)

From the second equation, we can isolate \( \cos(D) \):
\[ \cos(D) = \frac{1}{2\cos(S) \sqrt{2}} \]

### Step 3: Substitute \( \cos(D) \) into the first equation

Substituting \( \cos(D) \) into the first equation gives:
\[ 2 \sin(S) \cdot \frac{1}{2\cos(S) \sqrt{2}} = \sqrt{\frac{3}{2}} \]
This simplifies to:
\[ \frac{\sin(S)}{\cos(S) \sqrt{2}} = \sqrt{\frac{3}{2}} \]

### Step 4: Simplify and solve for \( \tan(S) \)

Cross-multiplying gives:
\[ \sin(S) = \sqrt{3} \cos(S) \]
This implies:
\[ \tan(S) = \sqrt{3} \]

### Step 5: Find the angle \( S \)

The angle \( S \) that satisfies \( \tan(S) = \sqrt{3} \) is:
\[ S = \frac{\pi}{3} \]

### Step 6: Find \( A + B \)

Since \( S = \frac{A + B}{2} \):
\[ A + B = 2S = \frac{2\pi}{3} \]

### Step 7: Use the cosine equation to find \( D \)

Now using the second equation:
\[ 2 \cos(S) \cos(D) = \frac{1}{\sqrt{2}} \]
Substituting \( S = \frac{\pi}{3} \):
\[ 2 \cdot \frac{1}{2} \cdot \cos(D) = \frac{1}{\sqrt{2}} \]
This simplifies to:
\[ \cos(D) = \frac{1}{\sqrt{2}} \]
Thus, \( D = \frac{\pi}{4} \).

### Step 8: Find \( A \) and \( B \)

Now we can find \( A \) and \( B \):
\[ A = S + D = \frac{\pi}{3} + \frac{\pi}{4} = \frac{4\pi + 3\pi}{12} = \frac{7\pi}{12} \]
\[ B = S - D = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]

### Final Answer

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