Trigonometry & Inverse Trigonometry
Sine Rule Applications
Grade 11
Question:
<p>If inside triangle ABC, a, b, c and angle A are given and \(c\sin A < a < c\), then \(b_1\) and \(b_2\) are values of b</p>
<p>(a) \(b_1 + b_2 = 2c\cos A\)</p>
<p>(b) \(b_1 + b_2 = c\cos A\)</p>
Step-by-Step Solution
Key Concept: Recognize the two-solution case and use sum-to-product formulas with sine rule to find the relationship.
<p>By sine rule: \(\frac{a}{\sin A} = \frac{c}{\sin C}\), so \(\sin C = \frac{c\sin A}{a}\)</p><p>Since \(c\sin A < a < c\), we have \(\sin C < 1\), giving two possible values of C: \(C_1\) and \(C_2 = 180° - C_1\).</p><p>Then \(B_1 = 180° - A - C_1\) and \(B_2 = 180° - A - C_2\)</p><p>\(B_1 + B_2 = 360° - 2A - C_1 - C_2 = 360° - 2A - 180° = 180° - 2A\)</p><p>By sine rule: \(b_1 + b_2 = \frac{a(\sin B_1 + \sin B_2)}{\sin A} = 2c\cos A\)</p>
Correct Answer: A