If twice the square on the diameter of a circle is equal to sum of the squares on the sides of the inscribed triangle ABC, then `sin^2 A+sin^2 B+sin^2 C` is equal to

Options

A. 2
B. 3
C. 4
D. 1

Answer

To solve the problem, we need to establish the relationship between the diameter of the circle and the sides of the inscribed triangle ABC. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that twice the square of the diameter of a circle is equal to the sum of the squares of the sides of the inscribed triangle ABC. The diameter \(D\) of the circle is related to its radius \(R\) by the equation:
\[ D = 2R \]
Therefore, the square of the diameter is:
\[ D^2 = (2R)^2 = 4R^2 \]
Thus, twice the square of the diameter is:
\[ 2D^2 = 2 \times 4R^2 = 8R^2 \]

2. **Expressing the Sides of the Triangle**:
Let the sides of triangle ABC be denoted as \(a\), \(b\), and \(c\). According to the problem, we have:
\[ 8R^2 = a^2 + b^2 + c^2 \]

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

4. **Substituting into the Equation**:
Now, substituting \(a\), \(b\), and \(c\) into the equation \(a^2 + b^2 + c^2\):
\[ a^2 = (2R \sin A)^2 = 4R^2 \sin^2 A \]
\[ b^2 = (2R \sin B)^2 = 4R^2 \sin^2 B \]
\[ c^2 = (2R \sin C)^2 = 4R^2 \sin^2 C \]
Therefore:
\[ a^2 + b^2 + c^2 = 4R^2 (\sin^2 A + \sin^2 B + \sin^2 C) \]

5. **Setting Up the Equation**:
We can now equate the two expressions:
\[ 8R^2 = 4R^2 (\sin^2 A + \sin^2 B + \sin^2 C) \]

6. **Simplifying the Equation**:
Dividing both sides by \(4R^2\) (assuming \(R \neq 0\)):
\[ 2 = \sin^2 A + \sin^2 B + \sin^2 C \]

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