In a triangle `ABC,a/b=2+sqrt(3)` and `/_C=60^(@)` then angle `A` and `B` is

Options

A. `/_A=105^(@)`
B. `/_B=15^(@)`
C. `/_A=115^(@)`
D. `/_ B=105^(@)`

Explanation

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

### Step-by-Step Solution:

1. **Use the Angle Sum Property of a Triangle**:
\[ A + B + C = 180^\circ \]
Given \( C = 60^\circ \):
\[ A + B + 60^\circ = 180^\circ \]
Therefore,
\[ A + B = 120^\circ \]

**Hint**: Remember that the sum of angles in a triangle is always \( 180^\circ \).

2. **Apply the Law of Sines**:
According to the Law of Sines:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
From the given ratio \( \frac{a}{b} = 2 + \sqrt{3} \), we can express this as:
\[ \frac{\sin A}{\sin B} = 2 + \sqrt{3} \]

**Hint**: The Law of Sines relates the sides of a triangle to the sines of its angles.

3. **Express \(\sin A\) in terms of \(\sin B\)**:
Let \( \sin A = (2 + \sqrt{3}) \sin B \).

**Hint**: Use the relationship between the sides and angles to express one sine in terms of the other.

4. **Use the Sine Addition Formula**:
Since \( A + B = 120^\circ \), we can use the sine of the sum:
\[ \sin(120^\circ - B) = \sin A \]
Therefore,
\[ \sin A = \sin(120^\circ - B) = \sin 120^\circ \cos B - \cos 120^\circ \sin B \]
Knowing \( \sin 120^\circ = \frac{\sqrt{3}}{2} \) and \( \cos 120^\circ = -\frac{1}{2} \), we have:
\[ \sin A = \frac{\sqrt{3}}{2} \cos B + \frac{1}{2} \sin B \]

**Hint**: Use trigonometric identities to express angles in terms of sine and cosine.

5. **Set the two expressions for \(\sin A\) equal**:
\[ (2 + \sqrt{3}) \sin B = \frac{\sqrt{3}}{2} \cos B + \frac{1}{2} \sin B \]
Rearranging gives:
\[ (2 + \sqrt{3} - \frac{1}{2}) \sin B = \frac{\sqrt{3}}{2} \cos B \]

**Hint**: Combine like terms and isolate \(\sin B\) and \(\cos B\).

6. **Substitute \(\sin B\) and \(\cos B\)**:
Using the identity \(\sin^2 B + \cos^2 B = 1\), substitute \(\cos B\) in terms of \(\sin B\) and solve for \(\sin B\).

**Hint**: Use Pythagorean identities to express one trigonometric function in terms of another.

7. **Solve the resulting equations**:
After substituting and simplifying, you will find values for \( A \) and \( B \):
\[ A = 105^\circ, \quad B = 15^\circ \]

**Hint**: Check your calculations to ensure that both angles satisfy the triangle's angle sum property.

### Final Answer:
Thus, the angles are:
\[ \boxed{A = 105^\circ, B = 15^\circ} \]