Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по геометрии 8 класс Атанасян ФГОС Задание 1049
Задание 1049
\[\boxed{\mathbf{1049.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[A\left( - 1;\sqrt{3} \right);\]
\[B\left( 1; - \sqrt{3} \right);\]
\[C\left( \frac{1}{2};\sqrt{3} \right).\]
\[\mathbf{Найти:}\]
\[\angle A;\angle B;\angle C.\]
\[\mathbf{Решение.}\]
\[1)\ AB =\]
\[= \sqrt{(1 + 1)^{2} + \left( \sqrt{3} + \sqrt{3} \right)^{2}} =\]
\[= \sqrt{4 + 12} = \sqrt{16} = 4;\]
\[BC =\]
\[= \sqrt{\left( 1 - \frac{1}{2} \right)^{2} + \left( - \sqrt{3} - \sqrt{3} \right)^{2}} =\]
\[= \sqrt{\frac{1}{4} + 12} = \sqrt{\frac{49}{4}} = 3,5;\]
\[AC = \sqrt{\left( \frac{1}{2} + 1 \right)^{2} + \left( \sqrt{3} - \sqrt{3} \right)^{2}} =\]
\[= \sqrt{\frac{9}{4}} = 1,5.\]
\[2)\ По\ теореме\ косинусов:\]
\[AB^{2} =\]
\[= BC^{2} + AC^{2} - 2BC \bullet AC \bullet \cos{\angle C}\]
\[16 = \frac{49}{4} + \frac{9}{4} - 2 \bullet \frac{7}{2} \bullet \frac{3}{2}\ \bullet \cos{\angle C}\]
\[16 = \frac{58}{4} - \frac{42}{4} \bullet \cos{\angle C}\]
\[\frac{42}{4} \bullet \cos{\angle A} = - \frac{6}{4}\]
\[\cos{\angle C} = - \frac{6}{4} \bullet \frac{4}{42} = - \frac{1}{7} \approx\]
\[\approx - 0,1429 < 0.\]
\[\angle C - тупой \Longrightarrow \angle C =\]
\[= 180{^\circ} - 81{^\circ}47^{'} = 98{^\circ}13^{'}.\]
\[BC^{2} =\]
\[= AB^{2} + AC^{2} - 2AB \bullet AC \bullet \cos{\angle A}\]
\[\frac{49}{4} = 16 + \frac{9}{4} - 2 \bullet 4 \bullet \frac{3}{2}\ \bullet \cos{\angle A}\]
\[\frac{40}{4} - 16 = - 12 \bullet \cos{\angle A}\]
\[- 6 = - 12\cos{\angle A}\]
\[\cos{\angle A} = \frac{6}{12} = \frac{1}{2}\ \]
\[\angle A = 60{^\circ}.\]
\[3)\ \angle B =\]
\[= 180{^\circ} - \left( 98{^\circ}13^{'} + 60{^\circ} \right) \approx\]
\[\approx 21{^\circ}47^{'}.\]
\[\mathbf{Ответ:}\angle A = 60{^\circ};\angle B = 21{^\circ}47^{'};\]
\[\angle C = 98{^\circ}13'\mathbf{.}\]
\[\boxed{\mathbf{1049.еуроки - ответы\ на\ пятёрку}}\]
\[x^{2} + y^{2} = 25\]
\[\textbf{а)}\ x = - 4:\]
\[( - 4)^{2} + y^{2} = 25;\]
\[16 + y^{2} = 25;\]
\[y^{2} = 9;\]
\[y = \pm 3.\]
\[A( - 4;3)\ и\ B( - 4; - 3).\]
\[\textbf{б)}\ y = 3:\]
\[x^{2} + 9 = 25;\]
\[x^{2} = 16;\]
\[x^{2} = \pm 4.\]
\[C(4;3)\ и\ D( - 4;3).\]
