Trigonometry & Inverse Trigonometry
Law of Cosines
Grade 11
Question:
<p>Sides of a triangle ABC are in AP. If \(a < \min\{b, c\}\), then \(\cos A\) may be equal to</p>
<p>(a) \(\frac{3c - 4b}{2b}\)</p>
<p>(b) \(\frac{3c - 4b}{2c}\)</p>
<p>(c) \(\frac{4c - 3b}{2b}\)</p>
<p>(d) \(\frac{4c - 3b}{2c}\)</p>
Step-by-Step Solution
Key Concept: Express sides in A.P. form and apply the law of cosines to find \(\cos A\).
<p>Given: Sides in A.P. and \(a < \min\{b, c\}\).</p><p>Let sides be \(b-d, b, b+d\) where \(a = b-d\), \(c = b+d\).</p><p>Using law of cosines: \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)</p><p>Substitute: \(\cos A = \frac{b^2 + (b+d)^2 - (b-d)^2}{2b(b+d)} = \frac{3b^2 + 4bd}{2b(b+d)} = \frac{3c - 4b}{2b}\)</p>
Correct Answer: A