Trigonometry & Inverse Trigonometry
Properties of Triangle
Grade 11
Question:
<p>In a triangle \(ABC\), \(a = 4\), \(b = 3\), \(\angle A = 60°\), then \(c\) is the root of the equation</p>
<p>\(c^2 - 3c - 7 = 0\)</p>
<p>\(c^2 + 3c + 7 = 0\)</p>
<p>\(c^2 - 3c + 7 = 0\)</p>
<p>\(c^2 + 3c - 7 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the Law of Cosines with given a, b, and angle A to form a quadratic equation in c. The equation a² = b² + c² - 2bc·cos(A) directly yields the required relation.
<p><strong>Step 1:</strong> Apply the Law of Cosines: a² = b² + c² - 2bc·cos(A)</p><p><strong>Step 2:</strong> Substitute a = 4, b = 3, ∠A = 60°, cos(60°) = 1/2:</p><p>16 = 9 + c² - 2(3)(c)·(1/2)</p><p>16 = 9 + c² - 3c</p><p><strong>Step 3:</strong> Rearrange to standard form:</p><p>c² - 3c + 9 - 16 = 0</p><p>c² - 3c - 7 = 0</p><p>∴ Answer: A (c² - 3c - 7 = 0)</p>
Correct Answer: A