Решебник по геометрии 9 класс Мерзляк Задание 87

\[Дан\ \mathrm{\Delta}\text{ABC.}\]

\[\frac{\text{AC}}{\sin{\angle B}} = \frac{\text{BC}}{\sin{\angle A}}\]

\[\sin{\angle A} = \frac{BC \bullet \sin{\angle B}}{\text{AC}}.\]

\[1)\ AC = 2\ см;BC = 1\ см;\]

\[\angle B = 135{^\circ}:\]

\[\sin{\angle A} = \frac{1}{2} \bullet \sin{135{^\circ}}\]

\[\sin{\angle A} = \frac{1}{2} \bullet \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}\]

\[\angle A = \arcsin\frac{\sqrt{2}}{4} \approx 21{^\circ}.\]

\[2)\ AC = \sqrt{2}\ см;BC = \sqrt{3}\ см;\]

\[\angle B = 45{^\circ}:\]

\[\sin{\angle A} = \frac{\sqrt{3}}{\sqrt{2}} \bullet \sin{45{^\circ}}\]

\[\sin{\angle A} = \frac{\sqrt{3}}{\sqrt{2}} \bullet \frac{\sqrt{2}}{2} = \frac{\sqrt{3}}{2}\]

\[\angle A = \arcsin\frac{\sqrt{3}}{2} = 60{^\circ}\]

\[\angle A = 180{^\circ} - 60{^\circ} = 120{^\circ}.\]

\[Ответ:\ \ 1)\ \approx 21{^\circ};\ 2)\ 60{^\circ};\ 120{^\circ}.\]