Question:
If \( \cos^2 A + \cos^2 C = \sin^2 B \), then \( \triangle ABC \) is
If \( \cos^2 A + \cos^2 C = \sin^2 B \), then \( \triangle ABC \) is
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Remember identity: \( \cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos C = 1 \) in triangles.
Updated On: Apr 23, 2026
- equilateral
- right angled
- isosceles
- None of the above
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The Correct Option is B
Solution and Explanation
Concept:
Use identity:
\[
\sin^2 B = 1 - \cos^2 B
\]
Step 1: Substitute identity. \[ \cos^2 A + \cos^2 C = 1 - \cos^2 B \] \[ \Rightarrow \cos^2 A + \cos^2 B + \cos^2 C = 1 \]
Step 2: Use identity in triangle: \[ \cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos C = 1 \] Comparing: \[ 2\cos A \cos B \cos C = 0 \]
Step 3: Hence: \[ \cos A = 0 \; \text{or} \; \cos B = 0 \; \text{or} \; \cos C = 0 \] \[ \Rightarrow \text{one angle } = 90^\circ \] Final Answer: \[ \text{Right angled triangle} \]
Step 1: Substitute identity. \[ \cos^2 A + \cos^2 C = 1 - \cos^2 B \] \[ \Rightarrow \cos^2 A + \cos^2 B + \cos^2 C = 1 \]
Step 2: Use identity in triangle: \[ \cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos C = 1 \] Comparing: \[ 2\cos A \cos B \cos C = 0 \]
Step 3: Hence: \[ \cos A = 0 \; \text{or} \; \cos B = 0 \; \text{or} \; \cos C = 0 \] \[ \Rightarrow \text{one angle } = 90^\circ \] Final Answer: \[ \text{Right angled triangle} \]
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