<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 area formula [ABC] = (abc)/(4R) and the projection formula a·cosA + b·cosB + c·cosC = (a² + b² + c²)/(2R) to establish a relationship between the given condition and the required expression.
<p><strong>Step 1:</strong> Recall the projection formula in a triangle:</p><p>a·cosA + b·cosB + c·cosC = (a² + b² + c²)/(2R)</p><p><strong>Step 2:</strong> Use the extended sine rule and area formula. We know [ABC] = (abc)/(4R), so given [ABC]/R = 4, we have:</p><p>abc/(4R²) = 4 ⟹ abc = 16R²</p><p><strong>Step 3:</strong> Also, from the sine rule: a = 2R·sinA, b = 2R·sinB, c = 2R·sinC</p><p>Therefore: a² + b² + c² = 4R²(sin²A + sin²B + sin²C)</p><p><strong>Step 4:</strong> By the identity in triangles: sin²A + sin²B + sin²C = 2(1 + cosA·cosB·cosC)</p><p>However, use the direct relation: a·cosA + b·cosB + c·cosC relates to the semiperimeter and circumradius.</p><p><strong>Step 5:</strong> From [ABC] = (1/2)(a·b·sinC) and using a·cosA + b·cosB + c·cosC = 4R·(sin2A + sin2B + sin2C)/(2sinA·sinB·sinC)... </p><p>More directly: When [ABC]/R = 4, applying the formula a·cosA + b·cosB + c·cosC = 2·[ABC]/R</p><p>∴ a·cosA + b·cosB + c·cosC = 2 × 4 = <strong>8</strong></p>
Correct Answer: C