Let S = ∑_ k=1 ^ 5 ^2 ( (kπ)/(11) ) and P = _ k=1 ^ 5 ( (kπ)/(11) )...

Given, $S = \sum_{k=1}^{5} \sin^2 \left( \dfrac{k\pi}{11} \right)$

Using the identity $\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$, we get:
$S = \dfrac{1}{2} \sum_{k=1}^{5} \left( 1 - \cos \dfrac{2k\pi}{11} \right) = \dfrac{5}{2} - \dfrac{1}{2} \sum_{k=1}^{5} \cos \dfrac{2k\pi}{11}$

Let $C = \cos \dfrac{2\pi}{11} + \cos \dfrac{4\pi}{11} + \cos \dfrac{6\pi}{11} + \cos \dfrac{8\pi}{11} + \cos \dfrac{10\pi}{11}$
Multiplying and dividing by $2 \sin \dfrac{\pi}{11}$:
$C = \dfrac{2 \sin \dfrac{\pi}{11} \cos \dfrac{2\pi}{11} + \dots + 2 \sin \dfrac{\pi}{11} \cos \dfrac{10\pi}{11}}{2 \sin \dfrac{\pi}{11}}$

Using $2 \sin A \cos B = \sin(A+B) - \sin(B-A)$:
$C = \dfrac{(\sin \dfrac{3\pi}{11} - \sin \dfrac{\pi}{11}) + (\sin \dfrac{5\pi}{11} - \sin \dfrac{3\pi}{11}) + \dots + (\sin \dfrac{11\pi}{11} - \sin \dfrac{9\pi}{11})}{2 \sin \dfrac{\pi}{11}}$

This is a telescoping sum that simplifies to:
$C = \dfrac{\sin \pi - \sin \dfrac{\pi}{11}}{2 \sin \dfrac{\pi}{11}} = -\dfrac{1}{2}$

Thus, $S = \dfrac{5}{2} - \dfrac{1}{2} \left( -\dfrac{1}{2} \right) = \dfrac{11}{4}$

Now for the product $P = \cos \dfrac{\pi}{11} \cos \dfrac{2\pi}{11} \cos \dfrac{3\pi}{11} \cos \dfrac{4\pi}{11} \cos \dfrac{5\pi}{11}$
We know $\cos \dfrac{3\pi}{11} = -\cos \left( \pi - \dfrac{3\pi}{11} \right) = -\cos \dfrac{8\pi}{11}$
And $\cos \dfrac{5\pi}{11} = -\cos \left( \pi - \dfrac{5\pi}{11} \right) = -\cos \dfrac{6\pi}{11} = -\cos \left( 2\pi - \dfrac{16\pi}{11} \right) = -\cos \dfrac{16\pi}{11}$

Substituting these into $P$:
$P = \cos \dfrac{\pi}{11} \cos \dfrac{2\pi}{11} \left( -\cos \dfrac{8\pi}{11} \right) \cos \dfrac{4\pi}{11} \left( -\cos \dfrac{16\pi}{11} \right)$
$P = \cos \dfrac{\pi}{11} \cos \dfrac{2\pi}{11} \cos \dfrac{4\pi}{11} \cos \dfrac{8\pi}{11} \cos \dfrac{16\pi}{11}$

Using the identity $\prod_{k=0}^{n-1} \cos(2^k \theta) = \dfrac{\sin(2^n \theta)}{2^n \sin \theta}$ with $n=5$ and $\theta = \dfrac{\pi}{11}$:
$P = \dfrac{\sin \dfrac{32\pi}{11}}{2^5 \sin \dfrac{\pi}{11}} = \dfrac{\sin \left( 3\pi - \dfrac{\pi}{11} \right)}{32 \sin \dfrac{\pi}{11}} = \dfrac{\sin \dfrac{\pi}{11}}{32 \sin \dfrac{\pi}{11}} = \dfrac{1}{32}$

Finally, the ratio $\dfrac{S}{P} = \dfrac{\dfrac{11}{4}}{\dfrac{1}{32}} = \dfrac{11}{4} \times 32 = 88$.

Answer: $88$