\[Схематический\ рисунок.\]
\[Дано:\]
\[AB = 15\ см;\]
\[BC = 7\ см;\]
\[AC = 13\ см;\]
\[AK = 8\ см;\]
\[MC = 3\ см.\]
\[Найти:\]
\[\text{MK.}\]
\[Решение.\]
\[1)\ В\ \mathrm{\Delta}ABC:\]
\[169 = 225 + 49 - 2 \bullet 15 \bullet 7 \bullet \cos{\angle B}\]
\[210\cos{\angle B} = 105\ \ \ \]
\[\cos{\angle B} = \frac{1}{2}.\]
\[2)\ В\ \mathrm{\Delta}BKM:\]
\[BK = AB - AK = 15 - 8 = 7\ см.\]
\[BM = BC + CM = 7 + 3 = 10\ см.\]
\[= 49 + 100 - 2 \bullet 7 \bullet 10 \bullet \frac{1}{2} =\]
\[MK^{2} = 149 - 70 = 79\ \ \ \]
\[MK = \sqrt{79}\ см.\]
\[Ответ:\ \ \sqrt{79}\ см.\]