In a `Delta ABC a/b=2+sqrt(3) " and " angle C = 60 ^@` Then the ordered pair `(angle A , angle B)` is equal to :

Options

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

Answer

To solve the problem, we need to find the ordered pair \((\angle A, \angle B)\) in triangle \(ABC\) given that \(\frac{a}{b} = 2 + \sqrt{3}\) and \(\angle C = 60^\circ\).

### Step-by-Step Solution:

1. **Use the Law of Sines**: According to the Law of Sines, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
From this, we can express the ratio \(\frac{a}{b}\) as:
\[ \frac{a}{b} = \frac{\sin A}{\sin B} \]

2. **Substitute the Given Ratio**: We know that \(\frac{a}{b} = 2 + \sqrt{3}\). Therefore, we can write:
\[ \frac{\sin A}{\sin B} = 2 + \sqrt{3} \]

3. **Express \(\sin A\) in terms of \(\sin B\)**: Rearranging gives us:
\[ \sin A = (2 + \sqrt{3}) \sin B \]

4. **Use the Angle Sum Property**: Since the angles in a triangle sum up to \(180^\circ\), we have:
\[ A + B + C = 180^\circ \]
Substituting \(C = 60^\circ\):
\[ A + B + 60^\circ = 180^\circ \implies A + B = 120^\circ \]
Thus, we can express \(B\) in terms of \(A\):
\[ B = 120^\circ - A \]

5. **Substitute for \(\sin B\)**: Now substituting \(B\) into the equation for \(\sin A\):
\[ \sin A = (2 + \sqrt{3}) \sin(120^\circ - A) \]

6. **Use the Sine Difference Identity**: We can use the sine difference identity:
\[ \sin(120^\circ - A) = \sin 120^\circ \cos A - \cos 120^\circ \sin A \]
Knowing that \(\sin 120^\circ = \frac{\sqrt{3}}{2}\) and \(\cos 120^\circ = -\frac{1}{2}\), we substitute:
\[ \sin(120^\circ - A) = \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A \]

7. **Set Up the Equation**: Plugging this back into our equation gives:
\[ \sin A = (2 + \sqrt{3}) \left( \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A \right) \]

8. **Rearranging the Equation**: Rearranging this equation leads to:
\[ \sin A - (2 + \sqrt{3}) \cdot \frac{1}{2} \sin A = (2 + \sqrt{3}) \cdot \frac{\sqrt{3}}{2} \cos A \]
This simplifies to:
\[ \left(1 - \frac{2 + \sqrt{3}}{2}\right) \sin A = (2 + \sqrt{3}) \cdot \frac{\sqrt{3}}{2} \cos A \]

9. **Finding Angles**: After solving the above equation, we find that:
\[ A = 105^\circ \quad \text{and} \quad B = 15^\circ \]

10. **Conclusion**: Thus, the ordered pair \((\angle A, \angle B)\) is:
\[ (105^\circ, 15^\circ) \]

### Final Answer:
The ordered pair \((\angle A, \angle B)\) is \((105^\circ, 15^\circ)\).

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