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МатематикаГДЗ по геометрии 8 класс Атанасян Задание 1061
Задание 1061
\[\boxed{\mathbf{1061.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\textbf{а)}\ Дано:\]
\[AB = 5\ см;\]
\[AC = 7,5\ см;\ \]
\[\angle A = 135{^\circ}.\]
\[Найти:\]
\[BC;\ \]
\[\angle B\ и\ \angle C.\]
\[Решение.\]
\[1)\ По\ теореме\ косинусов:\]
\[BC^{2} =\]
\[= AB^{2} + AC^{2} - 2AB \bullet AC \bullet \cos{\angle A};\]
\[BC^{2} =\]
\[= 25 + 56,25 - 75\cos{135{^\circ}} =\]
\[= 81,25 + 75 \bullet 0,7071 =\]
\[= 134,2825;\]
\[BC = 11,59\ см.\]
\[2)\ AC^{2} =\]
\[= BC^{2} + AB^{2} - 2BC \bullet AB \bullet \cos{\angle B};\]
\[56,25 =\]
\[= 134,28 + 25 - 115,9 \bullet \cos{\angle B}\]
\[115,9\cos{\angle B} = 103,03\]
\[\cos{\angle B} = 0,88895\]
\[\angle B = 27{^\circ}15^{'}.\]
\[3)\ \angle C =\]
\[= 180{^\circ} - \left( 135{^\circ} + 27{^\circ}15^{'} \right) =\]
\[= 17{^\circ}45^{'}.\]
\[\textbf{б)}\ Дано:\]
\[AB = 2\sqrt{2}\ дм;\]
\[BC = 3\ дм;\ \]
\[\angle B = 45{^\circ}.\]
\[Найти:\]
\[AC;\ \]
\[\angle A\ и\ \angle C.\]
\[Решение.\]
\[1)\ По\ теореме\ косинусов:\]
\[AC^{2} =\]
\[= AB^{2} + BC^{2} - 2AB \bullet BC \bullet \cos{\angle B};\]
\[AC^{2} = 8 + 9 - 12\sqrt{2}\cos{45{^\circ}} =\]
\[= 17 - 12\sqrt{2} \bullet \frac{\sqrt{2}}{2} = 17 - 12 = 5;\]
\[AC = \sqrt{2}\ дм.\]
\[2)\ AB^{2} =\]
\[= AC^{2} + BC^{2} - 2AC \bullet BC \bullet \cos{\angle C};\]
\[8 = 5 + 9 - 6\sqrt{5} \bullet \cos{\angle C}\]
\[6\sqrt{5}\cos{\angle C} = 6\]
\[\cos{\angle C} = \frac{6}{6\sqrt{5}} = \frac{1}{\sqrt{5}} = 0,88895\]
\[\angle C = 63{^\circ}26^{'}.\]
\[3)\ \angle A = 180{^\circ} - \left( 63{^\circ}26^{'} + 45{^\circ} \right) =\]
\[= 71{^\circ}34^{'}.\]
\[\textbf{в)}\ Дано:\]
\[AC = 0,6\ м;\]
\[BC = \frac{\sqrt{3}}{4}\ дм;\ \]
\[\angle C = 150{^\circ}.\]
\[Найти:\]
\[AB;\ \]
\[\angle A\ и\ \angle B.\]
\[Решение.\]
\[1)\ AC = 0,6\ м = 6\ дм.\]
\[2)\ По\ теореме\ косинусов:\]
\[AB^{2} =\]
\[= AC^{2} + BC^{2} - 2AC \bullet BC \bullet \cos{\angle C};\]
\[AB^{2} =\]
\[= 36 + \frac{3}{16} - \frac{12\sqrt{3}}{4}\cos{150{^\circ}} =\]
\[= 36 + \frac{3}{16} - \frac{3\sqrt{3}}{2} = 40,96;\]
\[AB = 6,4\ дм.\]
\[3)\ AC^{2} =\]
\[= BC^{2} + AB^{2} - 2BC \bullet AB \bullet \cos{\angle B};\]
\[36 =\]
\[= \frac{3}{16} + 40,6875 - 2,64 \bullet \frac{\sqrt{3}}{4} \bullet \cos{\angle B}\]
\[4,875 = 5,5426 \bullet \cos{\angle B}\]
\[\cos{\angle B} = 0,8796\]
\[\angle B = 28{^\circ}24^{'}.\]
\[4)\ \angle A =\]
\[= 180{^\circ} - \left( 28{^\circ}24^{'} + 150{^\circ} \right) =\]
\(= 1{^\circ}36^{'}.\)
