\[\boxed{\mathbf{597.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - прямоугольный;\]
\[\angle C = 90{^\circ};\]
\[AC = b;\ \]
\[CB = b;\]
\[a = 12;\]
\[b = 15.\]
\[\mathbf{а)\ Выразить:}\]
\[\text{AB\ },tg\ \angle A,\ tg\ \angle B\]
\[через\ a\ и\text{\ b.}\]
\[\textbf{б)}\ Найти:\]
\[AB;\ \ tg\ \angle A;\ \ \ tg\ \angle B.\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ AB = \sqrt{a^{2} + b^{2}}\ \]
\[(по\ теореме\ Пифагора);\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{a}{b}\ \]
\[(по\ определению\ тангенса\ угла);\]
\[tg\ \angle B = \frac{\text{AC}}{\text{CB}} = \frac{b}{a}\ \]
\[(по\ определению\ тангенса\ угла).\]
\[\textbf{б)}\ AB = \sqrt{a^{2} + b^{2}} =\]
\[= \sqrt{144 + 225} = \sqrt{369} = 19,2.\]
\[tg\ \angle A = \frac{a}{b} = \frac{12}{15} \Longrightarrow \angle A = 38{^\circ}39^{'}.\]
\[tg\ \angle B = \frac{b}{a} = \frac{15}{12} \Longrightarrow \angle B = 51{^\circ}21^{'}.\]
\[\mathbf{Ответ:}\mathbf{а)}\ AB = \sqrt{a^{2} + b^{2}};\]
\[tg\ \angle A = \frac{a}{b};tg\ \angle B = \frac{b}{a};\]
\[\textbf{б)}\ AB = 19,2;\angle A = 38{^\circ}39^{'};\ \]
\[\angle B = 51{^\circ}21^{'}.\]
\[\boxed{\mathbf{597.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[\textbf{а)}\ AB = 24\ см;\]
\[BC = 25\ см;\]
\[AC = 7\ см.\]
\[\textbf{б)}\ AB = 15\ см;\]
\[BC = 17\ см;\]
\[AC = 8\ см\text{.\ }\]
\[\mathbf{Найти:}\]
\[наименьшую\ высоту.\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ По\ формуле\ Герона\ \]
\[\left( \ p = \frac{a + b + c}{2} \right):\]
\[S_{\text{ABC}} = \sqrt{p(p - a)(p - b)(p - c)};\]
\[p = \frac{24 + 25 + 7}{2} = 28\ см.\]
\[= \sqrt{28 \bullet 4 \bullet 3 \bullet 21} =\]
\[= \sqrt{7 \bullet 4 \bullet 4 \bullet 3 \bullet 3 \bullet 7} = 3 \bullet 4 \bullet 7 =\]
\[= 84\ см^{2}.\]
\[3)\ S_{\text{ABC}} = \frac{1}{2} \bullet BH \bullet AC = 84\]
\[\frac{1}{2}BH \bullet 7 = 84;\]
\[BH = \frac{84 \bullet 2}{7} = 24\ см.\]
\[S_{\text{ABC}} = \frac{1}{2} \bullet CF \bullet AB = 84\]
\[\frac{1}{2}CF \bullet 24 = 84\]
\[CF = \frac{84 \bullet 2}{24} = 7\ см.\]
\[S_{\text{ABC}} = \frac{1}{2} \bullet AE \bullet BC = 84\]
\[\frac{1}{2}AE \bullet 25 = 84\]
\[AE = \frac{84 \bullet 2}{25} = \frac{168}{25} = 6,72\ см.\]
\[\mathbf{Ответ}:AE = 6,72\ см.\]
\[\textbf{б)}\ 1)\ По\ формуле\ Герона\ \]
\[\left( = \frac{a + b + c}{2} \right):\]
\[S_{\text{ABC}} = \sqrt{p(p - a)(p - b)(p - c)};\]
\[p = \frac{15 + 17 + 8}{2} = 20\ см.\]
\[= \sqrt{20 \bullet 5 \bullet 3 \bullet 12} =\]
\[= \sqrt{4 \bullet 5 \bullet 5 \bullet 3 \bullet 3 \bullet 4} = 5 \bullet 4 \bullet 3 =\]
\[= 60\ см^{2}.\]
\[3)\ S_{\text{ABC}} = \frac{1}{2} \bullet BH \bullet AC = 60\]
\[\frac{1}{2}BH \bullet 8 = 60\]
\[BH = \frac{60 \bullet 2}{8} = 15\ см.\]
\[S_{\text{ABC}} = \frac{1}{2} \bullet CF \bullet AB = 60\]
\[\frac{1}{2}CF \bullet 15 = 60\]
\[CF = \frac{60 \bullet 2}{15} = 8\ см.\]
\[S_{\text{ABC}} = \frac{1}{2} \bullet AE \bullet BC = 60\]
\[\frac{1}{2}AE \bullet 17 = 60\]
\[AE = \frac{60 \bullet 2}{17} = \frac{120}{17} = 7\frac{1}{17}\ см.\]
\[\mathbf{Ответ}:AE = 7\frac{1}{17}\mathbf{.}\]