Question

If \( \cos^2 A + \cos^2 C = \sin^2 B \), then \( \triangle ABC \) is

• A. equilateral
• B. right angled
• C. isosceles
• D. None of the above

Hint:

Remember identity: \( \cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos C = 1 \) in triangles.

Answer 1

The Correct Option is B

To determine the type of triangle \( \triangle ABC \) given the condition \( \cos^2 A + \cos^2 C = \sin^2 B \), we can use trigonometric identities and properties of triangles.

1. Recall the Pythagorean identity: \(\sin^2\theta + \cos^2\theta = 1\). For angle B, we have: \(\cos^2 B = 1 - \sin^2 B\).
2. From the given condition: \(\cos^2 A + \cos^2 C = \sin^2 B\), we can use the identity for B: \(\cos^2 B = 1 - \sin^2 B\), giving us: \(\cos^2 A + \cos^2 C = 1 - \cos^2 B\).
3. In a triangle, the sum of angles is \( 180^\circ \), i.e., \(A + B + C = 180^\circ\). This implies: \(\cos C = \sin B\), because if \(A + C = 90^\circ\), then \(C = 90^\circ - A\)and \(\cos C = \sin(90^\circ - A) = \sin A\).
4. Thus, if \(\cos A + \cos C = 1\), it suggests that \(\triangle ABC\)is right-angled, specifically right-angled at B.

Therefore, based on the given condition, the triangle \( \triangle ABC \) is indeed a right-angled triangle. This matches the correct answer choice: right angled.