`Delta ABC` has a right angle at A. If BC = `sqrt2 and AB = AC = 1`, then sin B is equal to

Options

A. `(1)/(sqrt2)`
B. `sqrt2`
C. `(sqrt3)/(2)`
D. `sqrt3`

Answer

To solve the problem, we need to find the value of sin B in triangle ABC, where angle A is a right angle, AB = AC = 1, and BC = √2.

### Step-by-Step Solution:

1. **Understand the Triangle Configuration**:
- We have a right triangle ABC with right angle at A.
- Given that AB = AC = 1, we can denote the lengths of sides as follows:
- AB = 1 (one leg)
- AC = 1 (the other leg)
- BC = √2 (the hypotenuse)

2. **Identify the Sides**:
- In triangle ABC:
- Perpendicular side (opposite to angle B) = AC = 1
- Hypotenuse = BC = √2

3. **Use the Sine Definition**:
- The sine of angle B (sin B) is defined as the ratio of the length of the side opposite to angle B to the length of the hypotenuse.
- Therefore, we can write:
\[ \sin B = \frac{\text{Opposite side (AC)}}{\text{Hypotenuse (BC)}} \]

4. **Substitute the Known Values**:
- Substitute the values we have:
\[ \sin B = \frac{AC}{BC} = \frac{1}{\sqrt{2}} \]

5. **Simplify the Expression**:
- To express sin B in a more standard form, we can rationalize the denominator:
\[ \sin B = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

6. **Conclusion**:
- Thus, the value of sin B is:
\[ \sin B = \frac{\sqrt{2}}{2} \]