In a right angled triangle ABC sin^(2)A+sin^(2)B+sin^(2)C=

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we need to find the value of \( \sin^2 A + \sin^2 B + \sin^2 C \) in a right-angled triangle ABC, where angle C is the right angle.

### Step-by-Step Solution:

1. **Identify the Right Angle**:
In triangle ABC, let angle C be the right angle. Therefore, we have:
\[ C = 90^\circ \]

2. **Use the Pythagorean Identity**:
We know that in any triangle, the sum of angles is \( 180^\circ \). Thus, we can express the relationship between angles A and B:
\[ A + B + C = 180^\circ \implies A + B = 90^\circ \]

3. **Express Sine of Angle B**:
Since \( B = 90^\circ - A \), we can use the co-function identity for sine:
\[ \sin B = \sin(90^\circ - A) = \cos A \]

4. **Substitute into the Equation**:
Now we substitute \( \sin B \) into the original equation:
\[ \sin^2 A + \sin^2 B + \sin^2 C = \sin^2 A + \cos^2 A + \sin^2 90^\circ \]

5. **Calculate \( \sin^2 90^\circ \)**:
We know that:
\[ \sin 90^\circ = 1 \implies \sin^2 90^\circ = 1^2 = 1 \]

6. **Combine the Terms**:
Now, we can rewrite the equation:
\[ \sin^2 A + \cos^2 A + 1 \]

7. **Apply the Pythagorean Identity**:
From the Pythagorean identity, we know:
\[ \sin^2 A + \cos^2 A = 1 \]
Therefore, substituting this into our equation gives:
\[ 1 + 1 = 2 \]

8. **Final Result**:
Thus, we find that:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 2 \]

### Conclusion:
The value of \( \sin^2 A + \sin^2 B + \sin^2 C \) in a right-angled triangle ABC is \( 2 \).