\[\boxed{\mathbf{1031.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{а})\ a = 5;b = c = 4:\]
\[= b^{2} + c^{2} - 2bc \bullet cos\angle A\]
\[25 = 16 + 16 - 2 \bullet 16 \bullet cos\angle A\]
\[32cos\angle A = 32 - 25\]
\[32cos\angle A = 7\]
\[cos\angle A = \frac{7}{32} \approx 0,2188;\]
\[\angle A = 77{^\circ}22^{'}.\]
\[Напртив\ большей\ стороны\ \]
\[лежит\ больший\ угол:\]
\[\ \mathrm{\Delta}ABC - остроугольный.\]
\[\mathbf{б})\ a = 17;b = 8;с = 15:\]
\[a^{2} = b^{2} + c^{2} - 2bc \bullet cos\angle A\]
\[289 = 64 + 225 - 240 \bullet cos\angle A\]
\[240cos\angle A = 0\]
\[cos\angle A = 0\]
\[\angle A = 90{^\circ}.\]
\[\mathrm{\Delta}ABC - прямоугольный.\]
\[\mathbf{в})\ a = 9;b = 5;c = 6:\]
\[a^{2} = b^{2} + c^{2} - 2bc \bullet cos\angle A\]
\[81 = 52 + 36 - 60 \bullet cos\angle A\]
\[60cos\angle A = - 20\]
\[cos\angle A = - \frac{1}{3} < 0\]
\[\angle A - тупой.\]
\[\mathrm{\Delta}ABC - тупоугольный.\]
\[\boxed{\mathbf{1031.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[OBCA - трапеция;\]
\[OA = a;\]
\[BC = d;\]
\[A \in OX + ;\]
\[B(b;c).\]
\[\mathbf{Найти:}\]
\[\text{AC\ }и\ \text{OC.}\]
\[\mathbf{Решение.}\]
\[1)\ y_{C} = y_{B} =\]
\[= c\ (так\ как\ BC|\left| \text{OA} \right);\ \ \]
\[x_{C} = b + d:\]
\[C(b + d;c).\]
\[2)\ AC =\]
\[= \sqrt{(b + d - a)^{2} + (c - 0)^{2}} =\]
\[= \sqrt{(b + d - a)^{2} + c^{2}}.\]
\[3)\ OC =\]
\[= \sqrt{(b + d - 0)^{2} + (c - 0)^{2}} =\]
\[= \sqrt{{(b + d)}^{2} + c^{2}}.\]
\[\mathbf{Ответ:}\mathbf{\ }\]
\[AC = \sqrt{(b + d - a)^{2} + c^{2}};\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ CO = \sqrt{{(b + d)}^{2} + c^{2}}\mathbf{.}\]