4. В треугольнике KSP: SP = 12, ∠K = 60°, ∠P = 45°, KS = y, KP = x. Найти x и y.

∠S = 180° − 60° − 45° = 75°.

По теореме синусов: \(\dfrac{12}{\sin 60^\circ}=\dfrac{x}{\sin 75^\circ}=\dfrac{y}{\sin 45^\circ}\).

\(y=\dfrac{12\sin 45^\circ}{\sin 60^\circ}=4\sqrt{6}\).

\(x=\dfrac{12\sin 75^\circ}{\sin 60^\circ}=2\sqrt{3}(\sqrt{6}+\sqrt{2})\).

Ответ: \(x=2\sqrt{3}(\sqrt{6}+\sqrt{2})\), \(y=4\sqrt{6}\).