Step 1: Understanding the Concept:
We use trigonometric identities to simplify the given relation between the angles of a triangle (\(A+B+C = \pi\)).
Step 2: Key Formula or Approach:
1. \(\sin^{2} \theta = \frac{1 - \cos 2\theta}{2}\)
2. \(\cos X + \cos Y = 2 \cos \frac{X+Y}{2} \cos \frac{X-Y}{2}\)
: Detailed Explanation:
Given: \(\sin^{2} A + \sin^{2} B + \sin^{2} C = 2\)
\[ \frac{1 - \cos 2A}{2} + \frac{1 - \cos 2B}{2} + \sin^{2} C = 2 \]
\[ 1 - \frac{1}{2}(\cos 2A + \cos 2B) + \sin^{2} C = 2 \]
\[ \sin^{2} C - \frac{1}{2} [2 \cos(A+B) \cos(A-B)] = 1 \]
Since \(A+B = \pi - C\), \(\cos(A+B) = -\cos C\):
\[ (1 - \cos^{2} C) + \cos C \cos(A-B) = 1 \]
\[ \cos C \cos(A-B) - \cos^{2} C = 0 \]
\[ \cos C [\cos(A-B) - \cos C] = 0 \]
Using \(C = \pi - (A+B)\), so \(\cos C = -\cos(A+B)\):
\[ \cos C [\cos(A-B) + \cos(A+B)] = 0 \]
\[ \cos C [2 \cos A \cos B] = 0 \]
\[ 2 \cos A \cos B \cos C = 0 \]
This implies at least one of \(\cos A, \cos B, \text{ or } \cos C\) must be zero.
If \(\cos \theta = 0\) for an angle in a triangle, that angle is \(90^{\circ}\).
Step 3: Final Answer:
The triangle is right angled.