<p>The sides of a triangle are \(\sin\alpha\), \(\cos\alpha\) and \(\sqrt{1 + \sin\alpha\cos\alpha}\) for some \(0 < \alpha < \dfrac{\pi}{2}\). Then the greatest angle of the triangle is</p>
Step-by-Step Solution
Key Concept: For these three expressions to form a valid triangle, they must satisfy the triangle inequality. The critical insight is squaring the triangle inequality involving the longest side to avoid complex radical algebra and discovering which parameter constraint ensures all three inequalities hold simultaneously.
<p><strong>Step 1: Establish positivity constraint</strong></p><p>For these to be valid side lengths, we need sin α > 0 and cos α > 0, which means α ∈ (0, π/2).</p><p><strong>Step 2: Identify the longest side</strong></p><p>Since sin α, cos α ∈ (0,1) for α ∈ (0, π/2), we have sin²α + cos²α = 1, so sin α cos α < 1.</p><p>Thus √(1 + sin α cos α) > 1 > sin α, cos α, making it the longest side.</p><p><strong>Step 3: Apply critical triangle inequality</strong></p><p>The binding constraint is: sin α + cos α > √(1 + sin α cos α)</p><p>Squaring both sides (valid since both are positive):</p><p>(sin α + cos α)² > 1 + sin α cos α</p><p>sin²α + 2sin α cos α + cos²α > 1 + sin α cos α</p><p>1 + 2sin α cos α > 1 + sin α cos α</p><p>sin α cos α > 0 ✓ (automatically satisfied for α ∈ (0, π/2))</p><p><strong>Step 4: Check the other inequalities</strong></p><p>sin α + √(1 + sin α cos α) > cos α and cos α + √(1 + sin α cos α) > sin α are automatically satisfied since √(1 + sin α cos α) > 1.</p><p><strong>Step 5: Verify boundary behavior</strong></p><p>At α = π/4: sides are 1/√2, 1/√2, √(3/2). Triangle inequality is satisfied.</p><p>The triangle exists for all α ∈ (0, π/2).</p><p>∴ Answer: <strong>D</strong></p>
Correct Answer: D