If twice the square of the diameter of the circle is equal to half the sum of the squares of the sides of incribed triangle ABC,then `sin^(2)A+sin^(2)B+sin^(2)C` is equal to

Options

A. 1
B. 2
C. 4
D. 8

Answer

To solve the problem step by step, we will start from the given condition and work our way through the necessary calculations to find the value of \(\sin^2 A + \sin^2 B + \sin^2 C\).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that twice the square of the diameter of the circle is equal to half the sum of the squares of the sides of triangle ABC.
Mathematically, this can be expressed as:
\[ 2 \times (2R)^2 = \frac{1}{2}(a^2 + b^2 + c^2) \]

2. **Simplifying the Equation**:
The left side simplifies to:
\[ 2 \times 4R^2 = 8R^2 \]
So, we have:
\[ 8R^2 = \frac{1}{2}(a^2 + b^2 + c^2) \]

3. **Multiplying Both Sides by 2**:
To eliminate the fraction, multiply both sides by 2:
\[ 16R^2 = a^2 + b^2 + c^2 \]

4. **Using the Sine Rule**:
According to the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
From this, we can express the sides in terms of the angles:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

5. **Substituting Values of a, b, c**:
Substitute \(a\), \(b\), and \(c\) into the equation \(a^2 + b^2 + c^2 = 16R^2\):
\[ (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2 = 16R^2 \]

6. **Expanding the Equation**:
This expands to:
\[ 4R^2 \sin^2 A + 4R^2 \sin^2 B + 4R^2 \sin^2 C = 16R^2 \]

7. **Dividing by \(4R^2\)**:
Divide the entire equation by \(4R^2\):
\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{16R^2}{4R^2} \]

8. **Simplifying the Right Side**:
This simplifies to:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 4 \]

### Final Answer:
Thus, the value of \(\sin^2 A + \sin^2 B + \sin^2 C\) is \(4\).

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