If the radius of the circumcircle of the isosceles triangle ABC is equal to `AB(=AC),` then the angle A is equal to-

Options

A. `(pi)/(6)`
B. `(pi)/(3)`
C. `(pi)/(2)`
D. `(2pi)/(3)`

Answer

To solve the problem, we need to find the angle A of an isosceles triangle ABC where the radius of the circumcircle (R) is equal to the lengths of the equal sides (AB = AC).

### Step-by-Step Solution:

1. **Understanding the Triangle**:
We have an isosceles triangle ABC where AB = AC = r (the circumradius). Let the lengths of the sides be:
- AB = AC = r
- BC = a (the base)

2. **Using the Sine Rule**:
According to the sine rule in triangle ABC, we have:
\[ \frac{a}{\sin A} = 2R \]
Since R = r, we can rewrite it as:
\[ \frac{a}{\sin A} = 2r \]

3. **Finding the Angles B and C**:
Since the triangle is isosceles, angles B and C are equal. Let:
\[ B = C = x \]
Therefore, we have:
\[ A + 2x = 180^\circ \quad \text{(sum of angles in a triangle)} \]
This implies:
\[ A = 180^\circ - 2x \]

4. **Using the Sine Rule for Angles B and C**:
Applying the sine rule for angles B and C:
\[ \frac{r}{\sin B} = 2R \quad \text{and} \quad \frac{r}{\sin C} = 2R \]
Thus:
\[ \frac{r}{\sin x} = 2r \]
Simplifying gives:
\[ \sin x = \frac{1}{2} \]

5. **Finding the Value of x**:
The angle x for which \(\sin x = \frac{1}{2}\) is:
\[ x = 30^\circ \]

6. **Finding Angle A**:
Now substituting back to find angle A:
\[ A = 180^\circ - 2x = 180^\circ - 2(30^\circ) = 180^\circ - 60^\circ = 120^\circ \]

### Conclusion:
The angle A is equal to \(120^\circ\).