Trigonometry & Inverse Trigonometry
Cosine rule
Grade 11
Question:
<p>In a triangle, \(\cos A = \dfrac{b^2+c^2-a^2}{2bc}\). If \(a = 4\), \(b = 3\), \(\cos A = \cos 60°\), find \(c\).</p>
<p>\(c = 7\) or \(c = -1\)</p>
<p>\(c = 7\)</p>
<p>\(c^2 - 3c - 7 = 0\)</p>
<p>\(c = 3\)</p>
Step-by-Step Solution
Key Concept: Recognize that the given cosine formula is the Law of Cosines rearranged. Since cos A = 1/2 (from cos 60°), substitute directly into the Law of Cosines formula to find c using the quadratic equation.
<p><strong>Step 1:</strong> Identify that cos A = cos 60° = 1/2</p><p><strong>Step 2:</strong> Use the Law of Cosines formula given: cos A = (b² + c² - a²)/(2bc)</p><p><strong>Step 3:</strong> Substitute known values:</p><p>1/2 = (3² + c² - 4²)/(2·3·c)</p><p>1/2 = (9 + c² - 16)/(6c)</p><p>1/2 = (c² - 7)/(6c)</p><p><strong>Step 4:</strong> Cross-multiply:</p><p>6c · (1/2) = c² - 7</p><p>3c = c² - 7</p><p>c² - 3c - 7 = 0</p><p><strong>Step 5:</strong> Apply quadratic formula:</p><p>c = (3 ± √(9 + 28))/2 = (3 ± √37)/2</p><p><strong>Step 6:</strong> Since c must be positive (side length), reject the negative root:</p><p>c = (3 + √37)/2 ≈ 3.54</p><p>∴ Answer: C</p>
Correct Answer: C