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МатематикаГДЗ по геометрии 9 класс Атанасян Задание 1058
Задание 1058
\[\boxed{\mathbf{1058.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{а)\ }\]
\(\mathbf{\ }\)
\[\mathbf{б)}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[\textbf{а)}\ BC = 4,125\ м;\]
\[\angle C = 72{^\circ};\]
\[\textbf{б)}\ BC = 4100\ м;\]
\(\angle A = 32{^\circ};\)
\[\angle C = 120{^\circ}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[\mathbf{а)\ }1)\ \angle A = 180{^\circ} - (72{^\circ} + 44{^\circ}) =\]
\[= 64{^\circ}.\]
\[2)\ По\ теореме\ синусов:\]
\[\frac{\text{AB}}{\sin{\angle C}} = \frac{\text{BC}}{\sin{\angle A}} = \frac{\text{AC}}{\sin{\angle B}}\]
\[\frac{\text{AB}}{\sin{72{^\circ}}} = \frac{4,125}{\sin{64{^\circ}}};\]
\[AB = \frac{4,125 \bullet \sin{72{^\circ}}}{\sin{64{^\circ}}} \approx\]
\[\approx \frac{4,125 \bullet 0,9511}{0,8988} \approx 4,365\ м.\]
\[3)\ S_{\text{ABC}} = \frac{1}{2}AB \bullet BC \bullet \sin{\angle B} =\]
\[= \frac{1}{2} \bullet 4,125 \bullet 4,365 \bullet \sin{44{^\circ}} =\]
\[= \frac{1}{2} \bullet 4,125 \bullet 4,365 \bullet 0,695 \approx\]
\[\approx 6,25\ м^{2}.\]
\[\textbf{б)}\ 1)\ \angle A = 180{^\circ} - (120{^\circ} + 32{^\circ}) =\]
\[= 28{^\circ}.\]
\[2)\ По\ теореме\ синусов:\]
\[\frac{\text{AB}}{\sin{\angle C}} = \frac{\text{BC}}{\sin{\angle A}} = \frac{\text{AC}}{\sin{\angle B}}\]
\[\frac{\text{AB}}{\sin{120{^\circ}}} = \frac{4100}{\sin{32{^\circ}}};\]
\[AB = \frac{4100 \bullet \sin{120{^\circ}}}{\sin{32{^\circ}}} \approx\]
\[\approx \frac{4100 \bullet 0,866}{0,5299} \approx 6700,5\ м.\]
\[3)\ S_{\text{ABC}} = \frac{1}{2}AB \bullet BC \bullet \sin{\angle B} =\]
\[= \frac{1}{2} \bullet 4100 \bullet 6700,5\sin{28{^\circ}} =\]
\[= 0,5 \bullet 4100 \bullet 6700,5 \bullet 0,4695 \approx\]
\[\approx 6448673,1\ м^{2};\]
\[\mathbf{Ответ:\ а)}\ 6,25\mathbf{\ }м^{2}\mathbf{;}\]
\(\mathbf{б)\ }6448673,1\ м^{2}\mathbf{;}\mathbf{\ }\)
