Question

The sides of a triangle are \(\sin\alpha\), \(\cos\alpha\), and \(\sqrt{1+\sin\alpha\cos\alpha}\) for some \(0 < \alpha < \frac{\pi}{2}\). Then, the greatest angle of the triangle is:

• A. \(60^\circ\)
• B. \(90^\circ\)
• C. \(120^\circ\)
• D. \(150^\circ\)

Hint:

For triangle problems, always use the cosine rule to determine angles opposite the largest side.

Answer 1

The Correct Option is C

Step 1: Using the cosine rule. We are given the sides of the triangle as: \[ a = \sin\alpha, \quad b = \cos\alpha, \quad c = \sqrt{1 + \sin\alpha \cos\alpha}. \] To find the greatest angle, we use the cosine rule: \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab}. \] Substituting the values of \(a\), \(b\), and \(c\), we get: \[ \cos C = \frac{\sin^2\alpha + \cos^2\alpha - (1 + \sin\alpha \cos\alpha)}{2\sin\alpha \cos\alpha}. \] Using the identity \(\sin^2\alpha + \cos^2\alpha = 1\), this simplifies to: \[ \cos C = \frac{1 - (1 + \sin\alpha \cos\alpha)}{2\sin\alpha \cos\alpha} = \frac{-\sin\alpha \cos\alpha}{2\sin\alpha \cos\alpha} = -\frac{1}{2}. \] Therefore, \(C = 120^\circ\).