<p>In triangle \(ABC\) if \(\dfrac{[\Delta ABC]}{R} = 4\), then the value of \(a\cos A + b\cos B + c\cos C\) is:<br>[<strong>Note:</strong> \(R\) is the circumradius of triangle \(ABC\) and \([\Delta ABC]\) is the area of \(\Delta ABC\)]</p>
Step-by-Step Solution
Key Concept: Use the projection formula a·cos A + b·cos B + c·cos C = 4R·sin A·sin B·sin C, combined with the area formula [ABC] = 2R²·sin A·sin B·sin C to relate the given condition to the required expression.
<p><strong>Step 1:</strong> Use the projection formula. In any triangle: a·cos A + b·cos B + c·cos C = a·cos A + b·cos B + c·cos C. We can rewrite using the sine rule and projection properties.</p><p><strong>Step 2:</strong> By the extended sine rule: a = 2R·sin A, b = 2R·sin B, c = 2R·sin C</p><p>Therefore: a·cos A + b·cos B + c·cos C = 2R(sin A·cos A + sin B·cos B + sin C·cos C)</p><p><strong>Step 3:</strong> The area formula gives: [ABC] = (1/2)·ab·sin C = 2R²·sin A·sin B·sin C</p><p>Given [ABC]/R = 4, we have: 2R·sin A·sin B·sin C = 4, so sin A·sin B·sin C = 2/R</p><p><strong>Step 4:</strong> Using the identity in triangles: a·cos A + b·cos B + c·cos C = 4R·sin A·sin B·sin C</p><p>Substituting: a·cos A + b·cos B + c·cos C = 4R·(2/R) = 8</p><p>∴ Answer: C (which equals 8)</p>
Correct Answer: C