Question:

Step 1: Express the area of triangle ABC in terms of circumradius. The area of a triangle can be expressed as: $$[\triangle ABC] = \frac{abc}{4R}$$ Step 2: Use the given condition to find a relationship. Given that $\dfrac{[\triangle ABC]}{R} = 4$, we substitute the area formula: $$\frac{[\triangle ABC]}{R} = \frac{abc}{4R^2} = 4$$ This gives us: $$abc = 16R^2$$ Step 3: Express the sides in terms of circumradius and angles. Using the extended sine rule, we have: $$a = 2R\sin A, \quad b = 2R\sin B, \quad c = 2R\sin C$$ Step 4: Rewrite the target expression using the sine rule. We need to find $a\cos A + b\cos B + c\cos C$. Substituting the expressions from Step 3: $$a\cos A + b\cos B + c\cos C = 2R(\sin A\cos A + \sin B\cos B + \sin C\cos C)$$ Step 5: Apply the double angle formula. Using $2\sin\theta\cos\theta = \sin 2\theta$: $$a\cos A + b\cos B + c\cos C = R(\sin 2A + \sin 2B + \sin 2C)$$ Step 6: Use the identity for sum of sines of double angles in a triangle. For any triangle, there is a known identity: $$\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$$ Therefore: $$a\cos A + b\cos B + c\cos C = 4R\sin A\sin B\sin C$$ Step 7: Express the area in another form to find $\sin A\sin B\sin C$. The area of the triangle can also be written as: $$[\triangle ABC] = \frac{1}{2}ab\sin C = \frac{1}{2}(2R\sin A)(2R\sin B)\sin C = 2R^2\sin A\sin B\sin C$$ Step 8: Use the given condition to find $R\sin A\sin B\sin C$. From the given condition $\dfrac{[\triangle ABC]}{R} = 4$: $$\frac{2R^2\sin A\sin B\sin C}{R} = 4$$ $$2R\sin A\sin B\sin C = 4$$ $$R\sin A\sin B\sin C = 2$$ Step 9: Calculate the final answer. Substituting back into the expression from Step 6: $$a\cos A + b\cos B + c\cos C = 4R\sin A\sin B\sin C = 4 \times 2 = 8$$ **Final Answer: The value of $a\cos A + b\cos B + c\cos C$ is $\boxed{8}$, which corresponds to Option 3.**