Trigonometry & Inverse Trigonometry
Solution of Triangles
Grade 11
Question:
<p>Let a, b, c be sides of a triangle ABC and D denotes its area. If \(a = 2\), \(D = \sqrt{3}\), and \(a\cos C + \sqrt{3}a\sin C - b - c = 0\), then find the value of \((b + c)\).</p>
Step-by-Step Solution
Key Concept: Combine the trigonometric constraint with the area formula and use projection formulas to establish relationships between sides.
<p><strong>Solution:</strong> Given: \(a = 2\), \(D = \sqrt{3}\), and \(a\cos C + \sqrt{3}a\sin C - b - c = 0\).</p><p>From the third equation: \(2\cos C + 2\sqrt{3}\sin C = b + c\).</p><p>This can be rewritten as: \(4(\frac{1}{2}\cos C + \frac{\sqrt{3}}{2}\sin C) = 4\sin(C + 30°) = b + c\).</p><p>Using the area formula: \(D = \frac{1}{2}ab\sin C = \sqrt{3}\), so \(\sin C = \frac{\sqrt{3}}{b}\).</p><p>By the projection formula: \(b\cos C + c\cos B = a = 2\).</p><p>Solving simultaneously using the sine rule and the given constraint yields \(b + c = 4\).</p>
Correct Answer: 4