Trigonometry & Inverse Trigonometry
Properties of Triangle
Grade 11
Question:
<p>In a triangle \(ABC\), \(2ca\sin\dfrac{A-B+C}{2}\) is equal to</p>
<p>\(a^2 + b^2 - c^2\)</p>
<p>\(c^2 + a^2 - b^2\)</p>
<p>\(b^2 - c^2 - a^2\)</p>
<p>\(c^2 - a^2 - b^2\)</p>
Step-by-Step Solution
Key Concept: Use the angle relation A + B + C = π to simplify the argument of sine, then apply product-to-sum formulas or express in terms of sides using the sine rule.
<p><strong>Step 1:</strong> Simplify the angle using A + B + C = π.</p><p>A - B + C = (A + C) - B = (π - B) - B = π - 2B</p><p><strong>Step 2:</strong> Substitute into the expression.</p><p>2ca sin((π - 2B)/2) = 2ca sin(π/2 - B) = 2ca cos B</p><p><strong>Step 3:</strong> Apply the cosine rule: cos B = (a² + c² - b²)/(2ac)</p><p>2ca cos B = 2ca · (a² + c² - b²)/(2ac) = a² + c² - b²</p><p><strong>Step 4:</strong> Recognize this as the rearranged cosine rule.</p><p>By the cosine rule: b² = a² + c² - 2ac cos B, so a² + c² - b² = 2ac cos B</p><p>Alternatively, using sine rule and product-to-sum identities, this equals <strong>c² - b²</strong> or <strong>a² + c² - b²</strong> depending on the form expected.</p><p>∴ Answer: <strong>B</strong> (typically a² + c² - b² or equivalent form)</p>
Correct Answer: B