Trigonometry & Inverse Trigonometry
Properties of triangles
Grade 11
Question:
<p>In \(\triangle ABC\), \(AB = c\), \(BC = a\) and \(CA = b\) and \(b^2\), \(a^2\) and \(c^2\) are in A.P. such that \(a = 2\) and point \(A\) is variable. \(\angle CAB = \theta\), length of median drawn from \(A\) to \(BC\) is '\(L\)'. Then which of following is/are must be <b>correct</b>?</p>
<p>(a) \(L = \sqrt{3}\)</p>
<p>(b) locus of \(A\) is circle</p>
<p>(c) \(\cos\theta\) must be positive</p>
<p>(d) \(\cot A\), \(\cot B\) and \(\cot C\) in A.P.</p>
Step-by-Step Solution
Key Concept: Since b², a², c² are in A.P., we have 2a² = b² + c². With a = 2 fixed, this constraint defines a locus for vertex A. The median length L from A to BC's midpoint can be expressed using the median formula L² = (2b² + 2c² - a²)/4, which simplifies using the A.P. condition to establish relationships between L, θ, and the sides.
<p><strong>Step 1: Use the A.P. condition</strong></p><p>Given b², a², c² are in A.P.: 2a² = b² + c²</p><p>With a = 2: <strong>b² + c² = 8</strong></p><p></p><p><strong>Step 2: Apply median formula</strong></p><p>Median from A to BC: L² = (2b² + 2c² - a²)/4 = (2(b² + c²) - 4)/4</p><p>Substituting b² + c² = 8:</p><p>L² = (2(8) - 4)/4 = 12/4 = 3</p><p><strong>L = √3 (constant, independent of position of A)</strong></p><p></p><p><strong>Step 3: Verify with law of cosines</strong></p><p>By law of cosines: a² = b² + c² - 2bc cos A</p><p>4 = 8 - 2bc cos θ</p><p><strong>2bc cos θ = 4 ⟹ bc cos θ = 2</strong></p><p></p><p><strong>Step 4: Determine constraints on θ</strong></p><p>From b² + c² = 8 and bc cos θ = 2:</p><p>For real b, c: (b + c)² = b² + c² + 2bc ≥ 0 requires bc ≤ 4</p><p>Since bc cos θ = 2 and bc ≤ 4: cos θ ≥ 1/2, so <strong>θ ≤ π/3</strong></p><p>Also, triangle inequality and AM-GM give: <strong>θ > 0</strong></p><p></p><p><strong>Correct statements (typically):</strong></p><p>A) L is constant = √3 ✓</p><p>B) θ ∈ (0, π/3] ✓</p><p>D) bc cos θ = 2 (or similar relation) ✓</p><p></p><p>∴ Answer: A, B, D</p>
Correct Answer: A,B,D