\[\boxed{\mathbf{1043.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\left| \overrightarrow{P} \right| = 8;\]
\[\left| \overrightarrow{Q} \right| = 15;\]
\[\angle A = 120{^\circ}.\]
\[\mathbf{Найти:}\]
\[\left| \overrightarrow{F} \right| - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}PAA_{1} - прямоугольный:\]
\[\angle PAA_{1} = 120{^\circ} - 90{^\circ} = 30{^\circ}.\]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[PA_{1} = \frac{1}{2}AP = \frac{8}{2} = 4.\]
\[2)\ \left\{ \begin{matrix} AA_{1} = \sqrt{AP^{2} - A_{1}P^{2}} \\ AA_{1} = \sqrt{AF^{2} - A_{1}F^{2}} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \sqrt{AP^{2} - A_{1}P^{2}} =\]
\[= \sqrt{AF^{2} - A_{1}F^{2}};\]
\[8^{2} - 4^{2} = \overrightarrow{AF^{2}} - (15 - 4)\]
\[64 - 16 = \overrightarrow{AF^{2}} - 121\]
\[\overrightarrow{AF^{2}} = 64 + 121 - 16 = 169\]
\[\overrightarrow{\text{AF}} = \sqrt{169} = 13.\]
\[\mathbf{Ответ:}13\mathbf{.}\]