Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>The value of <span class="math">\(\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}\)</span> is</p>
<p>(a) <span class="math">\(\frac{a^2 + b^2 + c^2}{2abc}\)</span></p>
<p>(b) <span class="math">\(\frac{a^2 + b^2 + c^2}{abc}\)</span></p>
<p>(c) <span class="math">\(\frac{a^2 - b^2 + c^2}{abc}\)</span></p>
<p>(d) <span class="math">\(\frac{a^2 + b^2 - c^2}{abc}\)</span></p>
Step-by-Step Solution
Key Concept: Apply the cosine rule to each term, then add them to get the sum in terms of a, b, c.
<p>Using the cosine rule <span class="math">\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)</span> and similarly for <span class="math">\(\cos B\)</span> and <span class="math">\(\cos C\)</span>, we sum:</p><p><span class="math">\(\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2 + c^2 - a^2}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc} = \frac{a^2 + b^2 + c^2}{2abc}\)</span></p><p>∴ Answer is (a).</p>
Correct Answer: A