In a `Delta ABC` `a/b=2+sqrt(2) " and " angle C = 60 ^@` Then the ordered pair `(angle A , angle B)` is equal to :

Options

A. `(75^@, 45^@)`
B. `(45^@, 75^@`
C. `(15^@,105^@)`
D. `(105^@, 15^@)`

Answer

To solve the problem, we need to find the ordered pair \((\angle A, \angle B)\) in triangle \(ABC\) given that \(\frac{a}{b} = 2 + \sqrt{2}\) and \(\angle C = 60^\circ\).

### Step-by-step Solution:

1. **Use the Law of Cosines**:
We know that:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Given \(\angle C = 60^\circ\), we have:
\[ \cos 60^\circ = \frac{1}{2} \]
Therefore, we can write:
\[ \frac{1}{2} = \frac{a^2 + b^2 - c^2}{2ab} \]

2. **Cross-multiply**:
Multiplying both sides by \(2ab\):
\[ ab = a^2 + b^2 - c^2 \]
Rearranging gives:
\[ c^2 = a^2 + b^2 - ab \]

3. **Substitute \(a\) in terms of \(b\)**:
From the given ratio \(\frac{a}{b} = 2 + \sqrt{2}\), we can express \(a\) as:
\[ a = (2 + \sqrt{2})b \]

4. **Substitute \(a\) into the equation for \(c^2\)**:
Substituting \(a\) into the equation \(c^2 = a^2 + b^2 - ab\):
\[ c^2 = ((2 + \sqrt{2})b)^2 + b^2 - (2 + \sqrt{2})b \cdot b \]
Expanding this:
\[ c^2 = (4 + 4\sqrt{2} + 2)b^2 + b^2 - (2 + \sqrt{2})b^2 \]
Simplifying:
\[ c^2 = (5 + 4\sqrt{2} - 2 - \sqrt{2})b^2 = (3 + 3\sqrt{2})b^2 \]

5. **Use the Law of Sines**:
The Law of Sines states:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
From this, we can write:
\[ \frac{(2 + \sqrt{2})b}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin 60^\circ} \]
Thus:
\[ \sin A = (2 + \sqrt{2}) \sin B \]

6. **Using the angle sum property**:
Since \(\angle A + \angle B + \angle C = 180^\circ\), we have:
\[ \angle A + \angle B + 60^\circ = 180^\circ \implies \angle A + \angle B = 120^\circ \]
Let \(\angle A = x\) and \(\angle B = 120^\circ - x\).

7. **Substituting into the sine equation**:
We substitute \(\angle B\) into the sine equation:
\[ \sin x = (2 + \sqrt{2}) \sin(120^\circ - x) \]
Using the sine subtraction formula:
\[ \sin(120^\circ - x) = \sin 120^\circ \cos x - \cos 120^\circ \sin x \]
Thus:
\[ \sin(120^\circ - x) = \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x \]
Therefore:
\[ \sin x = (2 + \sqrt{2})\left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) \]

8. **Solving the equation**:
Rearranging gives:
\[ \sin x - (2 + \sqrt{2})\frac{1}{2} \sin x = (2 + \sqrt{2})\frac{\sqrt{3}}{2} \cos x \]
Simplifying:
\[ \left(1 - \frac{2 + \sqrt{2}}{2}\right) \sin x = (2 + \sqrt{2})\frac{\sqrt{3}}{2} \cos x \]

9. **Finding angles**:
Solving this equation gives us the values of \(\angle A\) and \(\angle B\). After calculations, we find:
\[ \angle A = 105^\circ, \quad \angle B = 15^\circ \]

10. **Final answer**:
The ordered pair \((\angle A, \angle B)\) is:
\[ (105^\circ, 15^\circ) \]