Trigonometry & Inverse Trigonometry
Properties of triangles
Grade 11
Question:
<p>The value of \(2ca\sin\!\left(\dfrac{A-B+C}{2}\right)\) equals</p>
<p>\(a^2 + c^2 - b^2\)</p>
<p>\(a^2 + b^2 - c^2\)</p>
<p>\(b^2 + c^2 - a^2\)</p>
<p>\(a^2 - b^2 - c^2\)</p>
Step-by-Step Solution
Key Concept: Use the angle sum property A + B + C = π to simplify the argument, then apply the product-to-sum formula with sine and cosine identities.
<p><strong>Step 1:</strong> Use the fundamental property of triangles: A + B + C = π</p><p><strong>Step 2:</strong> Simplify the argument:</p><p>A - B + C = (A + C) - B = (π - B) - B = π - 2B</p><p><strong>Step 3:</strong> Therefore:</p><p>sin((A - B + C)/2) = sin((π - 2B)/2) = sin(π/2 - B) = cos(B)</p><p><strong>Step 4:</strong> Substitute back:</p><p>2ca·sin((A - B + C)/2) = 2ca·cos(B)</p><p><strong>Step 5:</strong> Apply the law of cosines: cos(B) = (a² + c² - b²)/(2ac)</p><p>2ca·cos(B) = 2ca · (a² + c² - b²)/(2ac) = a² + c² - b²</p><p>∴ Answer: <strong>a² + c² - b²</strong> (Option A)</p>
Correct Answer: A