(i) To find angle BAC, use the cosine rule in triangle ABC:

\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)

where a = 16, b = 10, c = 10. Substituting the values:

\(\cos A = \frac{10^2 + 10^2 - 16^2}{2 \times 10 \times 10} = \frac{100 + 100 - 256}{200} = \frac{-56}{200} = -0.28\)

Thus,

\(A = \cos^{-1}(-0.28) \approx 0.6435 \text{ radians}\)

(ii) To find the area of the shaded region, calculate the area of triangle ABC and subtract it from the sum of the areas of the sectors:

Area of triangle ABC:

\(\text{Area} = \frac{1}{2} \times 16 \times 10 \times \sin(0.6435) = 48\)

Area of one sector (e.g., sector ABE):

\(\text{Area} = \frac{1}{2} \times 10^2 \times 0.6435 = 32.175\)

Total area of two sectors:

\(2 \times 32.175 = 64.35\)

Shaded area:

\(\text{Shaded area} = 64.35 - 48 = 16.35 \approx 16.4\)