\[Схематический\ рисунок.\]
\[Дано:\]
\[AB,\ AC - касательные;\]
\[O - центр\ окружности;\]
\[AB = AC = 12\ см;\]
\[BC = 14,4\ см.\]
\[Найти:\]
\[\text{OB.}\]
\[Решение:\]
\[1)\ OB\bot AB,\ \ \ \]
\[OC\bot AC,\ \ \ \]
\[OB = OC.\]
\[2)\ В\ \mathrm{\Delta}ABC:\]
\[\cos{\angle A} = \frac{AB^{2} + AC^{2} - BC^{2}}{2AB \bullet AC} =\]
\[= \frac{144 + 144 - 207,36}{2 \bullet 12 \bullet 12} =\]
\[= \frac{80,64}{288} = 0,28.\]
\[3)\ ABOC:\]
\[\angle O = 360{^\circ} - \angle A - \angle B - \angle C\]
\[\angle O = 180{^\circ} - \angle A;\]
\[\cos{\angle O} = - \cos{\angle A} = - 0,28.\]
\[4)\ В\ \mathrm{\Delta}BOC:\]
\[BC^{2} = OB^{2} + OC^{2} - 2OB \bullet OC\cos{\angle O}\]
\[OB^{2} + OB^{2} + 2OB \bullet OB \bullet 0,28 = 207,36\]
\[2OB^{2} + 0,56OB^{2} = 207,36\]
\[2,56OB^{2} = 207,36\ \ \]
\[OB^{2} = 81\ \ \ \]
\[OB = 9\ см.\]
\(Ответ:\ \ 9\ см.\)