Trigonometry & Inverse Trigonometry
Solution of Triangles
Grade 11
Question:
<p>The set of real numbers a such that \(a^2 + 2a\), \(2a + 3\), \(a^2 + 3a + 8\) are the sides of a triangle, is:</p>
<p>(a) \((0, \infty)\)</p>
<p>(b) \((5, 8)\)</p>
<p>(c) \(\left(-\frac{11}{3}, \infty\right)\)</p>
<p>(d) \((5, \infty)\)</p>
Step-by-Step Solution
Key Concept: For three lengths to form a triangle, they must be positive and satisfy the triangle inequality: the sum of any two sides must be greater than the third side. We need to find all values of a where all three conditions hold simultaneously.
<p><strong>Step 1: Ensure all sides are positive.</strong></p><p>Let p = a² + 2a, q = 2a + 3, r = a² + 3a + 8</p><p>For p > 0: a² + 2a > 0 ⟹ a(a + 2) > 0 ⟹ a < -2 or a > 0</p><p>For q > 0: 2a + 3 > 0 ⟹ a > -3/2</p><p>For r > 0: a² + 3a + 8 > 0. Discriminant = 9 - 32 = -23 < 0, so always positive.</p><p>Combined positivity: a > 0</p><p><strong>Step 2: Apply triangle inequality p + q > r.</strong></p><p>(a² + 2a) + (2a + 3) > a² + 3a + 8</p><p>a² + 4a + 3 > a² + 3a + 8</p><p>a > 5</p><p><strong>Step 3: Apply triangle inequality p + r > q.</strong></p><p>(a² + 2a) + (a² + 3a + 8) > 2a + 3</p><p>2a² + 5a + 8 > 2a + 3</p><p>2a² + 3a + 5 > 0</p><p>Discriminant = 9 - 40 = -31 < 0, so always true for all a.</p><p><strong>Step 4: Apply triangle inequality q + r > p.</strong></p><p>(2a + 3) + (a² + 3a + 8) > a² + 2a</p><p>a² + 5a + 11 > a² + 2a</p><p>3a + 11 > 0</p><p>a > -11/3</p><p><strong>Step 5: Find intersection of all conditions.</strong></p><p>From Step 1: a > 0</p><p>From Step 2: a > 5</p><p>From Step 3: always true</p><p>From Step 4: a > -11/3</p><p>The most restrictive condition is a > 5.</p><p><strong>∴ Answer: d</strong></p>
Correct Answer: d