8)

Решение:

В треугольнике ABC:

\( \angle BAC = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ} \).

В треугольнике ABK:

\( \angle ABK = 45^{\circ} \).

\( \angle AKB = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 75^{\circ} \).

Значит, треугольник ABK равнобедренный с \( AB = BK = 10 + 5 = 15 \).

По теореме синусов в треугольнике ABC:

\( \frac{AC}{\sin 45^{\circ}} = \frac{BC}{\sin 75^{\circ}} = \frac{AB}{\sin 60^{\circ}} \)

\( \frac{BC}{\sin 75^{\circ}} = \frac{15}{\sin 60^{\circ}} \)

\( BC = \frac{15 \cdot \sin 75^{\circ}}{\sin 60^{\circ}} = \frac{15 \cdot \frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{3}}{2}} = \frac{15(\sqrt{6} + \sqrt{2})}{4} \cdot \frac{2}{\sqrt{3}} = \frac{15(\sqrt{18} + \sqrt{6})}{6} = \frac{5(3\sqrt{2} + \sqrt{6})}{2} \).

В треугольнике BKC:

\( \angle BKC = 180^{\circ} - \angle AKB = 180^{\circ} - 75^{\circ} = 105^{\circ} \).

По теореме синусов в треугольнике BKC:

\( \frac{KC}{\sin \angle KBC} = \frac{BC}{\sin \angle BKC} \)

\( \frac{x}{\sin 45^{\circ}} = \frac{BC}{\sin 105^{\circ}} \)

\( x = \frac{BC \cdot \sin 45^{\circ}}{\sin 105^{\circ}} = \frac{\frac{5(3\sqrt{2} + \sqrt{6})}{2} \cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\frac{5(6 + \sqrt{12})}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{5(6 + 2\sqrt{3})}{\sqrt{6} + \sqrt{2}} = \frac{10(3 + \sqrt{3})}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{10(3\sqrt{6} - 3\sqrt{2} + \sqrt{18} - \sqrt{6})}{6-2} = \frac{10(2\sqrt{6} - 3\sqrt{2} + 3\sqrt{2})}{4} = \frac{20\sqrt{6}}{4} = 5\sqrt{6} \).

Ответ: \( 5\sqrt{6} \)