The angles `alpha , beta , gamma ` of a triangle satisfy the equations `2 sin alpha + 3 cos beta = 3sqrt(2)` and `3sin beta + 2cos alpha = 1`. Then angle `gamma` equals

Options

A. `150^(@)`
B. `120^(@)`
C. `60^(@)`
D. `30^(@)`

Answer

To solve the problem, we are given two equations involving the angles of a triangle, α, β, and γ:

1. \( 2 \sin \alpha + 3 \cos \beta = 3\sqrt{2} \) (Equation 1)
2. \( 3 \sin \beta + 2 \cos \alpha = 1 \) (Equation 2)

We need to find the value of angle γ.

### Step 1: Square both equations and add them
We start by squaring both equations and adding them together.

- Squaring Equation 1:
\[ (2 \sin \alpha + 3 \cos \beta)^2 = (3\sqrt{2})^2 \]
Expanding this, we get:
\[ 4 \sin^2 \alpha + 12 \sin \alpha \cos \beta + 9 \cos^2 \beta = 18 \]

- Squaring Equation 2:
\[ (3 \sin \beta + 2 \cos \alpha)^2 = 1^2 \]
Expanding this, we get:
\[ 9 \sin^2 \beta + 12 \sin \beta \cos \alpha + 4 \cos^2 \alpha = 1 \]

### Step 2: Combine the equations
Now, we add the two expanded equations:
\[ (4 \sin^2 \alpha + 9 \cos^2 \beta + 12 \sin \alpha \cos \beta) + (9 \sin^2 \beta + 4 \cos^2 \alpha + 12 \sin \beta \cos \alpha) = 18 + 1 \]
This simplifies to:
\[ 4 \sin^2 \alpha + 9 \sin^2 \beta + 4 \cos^2 \alpha + 9 \cos^2 \beta + 12 (\sin \alpha \cos \beta + \sin \beta \cos \alpha) = 19 \]

### Step 3: Use trigonometric identities
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 4(1 - \cos^2 \alpha) + 9(1 - \cos^2 \beta) + 12 \sin(\alpha + \beta) = 19 \]
This simplifies to:
\[ 4 + 9 - 4 \cos^2 \alpha - 9 \cos^2 \beta + 12 \sin(\alpha + \beta) = 19 \]
Thus:
\[ -4 \cos^2 \alpha - 9 \cos^2 \beta + 12 \sin(\alpha + \beta) = 6 \]

### Step 4: Solve for \( \sin(\alpha + \beta) \)
Rearranging gives:
\[ 12 \sin(\alpha + \beta) = 6 + 4 \cos^2 \alpha + 9 \cos^2 \beta \]
Now we can express \( \sin(\alpha + \beta) \) in terms of \( \alpha \) and \( \beta \).

### Step 5: Find \( \sin \gamma \)
Since \( \alpha + \beta + \gamma = 180^\circ \), we have:
\[ \sin(\alpha + \beta) = \sin(180^\circ - \gamma) = \sin \gamma \]
Thus:
\[ \sin \gamma = \frac{1}{2} \]

### Step 6: Determine \( \gamma \)
The angles for which \( \sin \gamma = \frac{1}{2} \) are:
\[ \gamma = 30^\circ \quad \text{or} \quad \gamma = 150^\circ \]
However, since \( \alpha + \beta + \gamma = 180^\circ \) and α and β must be positive angles, we conclude:
\[ \gamma = 30^\circ \]

### Final Answer
Thus, the value of angle γ is \( \boxed{30^\circ} \).