11. KL, \(\angle\) K, \(\angle\) L -?

Решение:

В треугольнике KML, \(\angle KML = 90^{\circ}\), \(\angle MKL = 30^{\circ}\), \( ML = 10 \).

В прямоугольном треугольнике KML:

\( \cdot\tan(30^{\circ}) = \frac{ML}{KM} \) \(\cdot\) \( KM = \frac{ML}{\tan(30^{\circ})} = \frac{10}{1/\sqrt{3}} = 10\cdot\sqrt{3} \).

\( \cdot\sin(30^{\circ}) = \frac{ML}{KL} \) \(\cdot\) \( KL = \frac{ML}{\sin(30^{\circ})} = \frac{10}{1/2} = 20 \).

\( \cdot\angle L = 90^{\circ} - \angle KML = 90^{\circ} - 30^{\circ} = 60^{\circ} \).

В треугольнике KLN:

\( \cdot\angle KNL = 90^{\circ} \), \( \cdot\angle L = 60^{\circ} \), \( \cdot\angle KLN = 30^{\circ} \).

Ответ: KL = 20.