ГДЗ по геометрии 9 класс Атанасян ФГОС Задание 1138

\[\boxed{\mathbf{1138.еуроки - ответы\ на\ пятёрку}}\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC;\]

\[A\left( - 1;\sqrt{3} \right);\]

\[B\left( 1; - \sqrt{3} \right);\]

\[C\left( \frac{1}{2};\sqrt{3} \right).\]

\[\mathbf{Найти:}\]

\[\angle A;\angle B;\angle C.\]

\[\mathbf{Решение.}\]

\[1)\ AB =\]

\[= \sqrt{(1 + 1)^{2} + \left( \sqrt{3} + \sqrt{3} \right)^{2}} =\]

\[= \sqrt{4 + 12} = \sqrt{16} = 4;\]

\[BC =\]

\[= \sqrt{\left( 1 - \frac{1}{2} \right)^{2} + \left( - \sqrt{3} - \sqrt{3} \right)^{2}} =\]

\[= \sqrt{\frac{1}{4} + 12} = \sqrt{\frac{49}{4}} = 3,5;\]

\[AC =\]

\[= \sqrt{\left( \frac{1}{2} + 1 \right)^{2} + \left( \sqrt{3} - \sqrt{3} \right)^{2}} =\]

\[= \sqrt{\frac{9}{4}} = 1,5.\]

\[2)\ По\ теореме\ косинусов:\]

\[16 = \frac{49}{4} + \frac{9}{4} - 2 \bullet \frac{7}{2} \bullet \frac{3}{2}\ \bullet \cos{\angle C}\]

\[16 = \frac{58}{4} - \frac{42}{4} \bullet \cos{\angle C}\]

\[\frac{42}{4} \bullet \cos{\angle A} = - \frac{6}{4}\]

\[\cos{\angle C} = - \frac{6}{4} \bullet \frac{4}{42} = - \frac{1}{7} \approx\]

\[\approx - 0,1429 < 0.\]

\[\angle C - тупой \Longrightarrow\]

\[\Longrightarrow \angle C = 180{^\circ} - 81{^\circ}47^{'} =\]

\[= 98{^\circ}13^{'}.\]

\[\frac{49}{4} = 16 + \frac{9}{4} - 2 \bullet 4 \bullet \frac{3}{2}\ \bullet \cos{\angle A}\]

\[\frac{40}{4} - 16 = - 12 \bullet \cos{\angle A}\]

\[- 6 = - 12\cos{\angle A}\]

\[\cos{\angle A} = \frac{6}{12} = \frac{1}{2}\ \]

\[\angle A = 60{^\circ}.\]

\[3)\ \angle B =\]

\[= 180{^\circ} - \left( 98{^\circ}13^{'} + 60{^\circ} \right) \approx\]

\[\approx 21{^\circ}47^{'}.\]

\[\mathbf{Ответ:}\angle A = 60{^\circ};\angle B = 21{^\circ}47^{'};\]

\[\angle C = 98{^\circ}13'\mathbf{.}\]