Find the area of an isoceles triangle ABC in which `angle ABC=120^@` and each of the smaller side is 1 unit.

Options

A. `sqrt3/4`
B. `sqrt3/2`
C. `2sqrt3`
D. `(3sqrt3)/10`

Answer

To find the area of the isosceles triangle ABC where angle ABC = 120° and each of the equal sides (AB and AC) is 1 unit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Triangle Configuration**:
- We have triangle ABC where AB = AC = 1 unit and angle ABC = 120°.

2. **Use the Area Formula for a Triangle**:
- The area \( A \) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times a \times b \times \sin(C) \]
where \( a \) and \( b \) are the lengths of two sides, and \( C \) is the included angle.

3. **Substitute the Known Values**:
- Here, \( a = AB = 1 \), \( b = AC = 1 \), and \( C = \angle ABC = 120° \).
- Plugging these values into the area formula gives:
\[ A = \frac{1}{2} \times 1 \times 1 \times \sin(120°) \]

4. **Calculate \( \sin(120°) \)**:
- We know that \( \sin(120°) = \sin(180° - 60°) = \sin(60°) = \frac{\sqrt{3}}{2} \).

5. **Calculate the Area**:
- Now substituting \( \sin(120°) \) into the area formula:
\[ A = \frac{1}{2} \times 1 \times 1 \times \frac{\sqrt{3}}{2} \]
- This simplifies to:
\[ A = \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \]

6. **Final Result**:
- Therefore, the area of triangle ABC is:
\[ A = \frac{\sqrt{3}}{4} \text{ square units} \]