Решебник по геометрии 9 класс Мерзляк Задание 115

\[Прямоугольный\ треугольник.\]

\[1)\ a = 7\ см,\ \ \ b = 35\ см:\]

\[c = \sqrt{a^{2} + b^{2}} = \sqrt{49 + 1225} =\]

\[= 35,7\ см;\]

\[\sin\alpha = \frac{a}{c} = \frac{7}{35,7} \approx 0,1961;\ \ \ \]

\[\alpha \approx 11{^\circ};\]

\[\gamma = 90{^\circ};\ \ \]

\[\beta = 90{^\circ} - 11{^\circ} = 79{^\circ}.\]

\[2)\ c = 17\ см,\ \ \ a = 8\ см:\]

\[b = \sqrt{c^{2} - a^{2}} = \sqrt{289 - 64} =\]

\[= 15\ см;\]

\[\sin\alpha = \frac{a}{c} = \frac{8}{17} \approx 0,4706;\ \ \]

\[\alpha \approx 28{^\circ};\]

\[\gamma = 90{^\circ};\ \ \ \]

\[\beta = 90{^\circ} - 28{^\circ} = 62{^\circ}.\]

\[3)\ c = 4\ см,\ \ \ \alpha = 50{^\circ}:\]

\[a = c \bullet \sin\alpha = 4 \bullet \sin{50{^\circ}} \approx 3,1\ см;\]

\[b = c \bullet \cos\alpha = 4 \bullet \cos{50{^\circ}} \approx 2,6\ см;\]

\[\gamma = 90{^\circ};\ \ \ \]

\[\beta = 90{^\circ} - 50{^\circ} = 40{^\circ}.\]

\[4)\ a = 8\ см,\ \ \ \alpha = 42{^\circ}:\]

\[c = \frac{a}{\sin\alpha} = \frac{8}{\sin{42{^\circ}}} \approx 12\ см;\]

\[b = \frac{a}{\text{tg\ α}} = \frac{8}{tg\ 42{^\circ}} \approx 8,9\ см;\]

\[\gamma = 90{^\circ};\ \ \]

\[\beta = 90{^\circ} - 42{^\circ} = 48{^\circ}.\]