Решебник по геометрии 9 класс Мерзляк Задание 124

\[1)\ a = 23\ см,\ c = 30\ см,\ \gamma = 102{^\circ}:\]

\[\sin\alpha = \frac{a\sin\gamma}{c} = \frac{23 \bullet \sin{102{^\circ}}}{30};\]

\[\sin\alpha \approx 0,7499;\ \ \ \]

\[\alpha \approx 49{^\circ};\]

\[\beta = 180{^\circ} - 102{^\circ} - 49{^\circ} = 29{^\circ};\]

\[b = \frac{c\sin\beta}{\sin\gamma} = \frac{30 \bullet \sin{29{^\circ}}}{\sin{102{^\circ}}} \approx 14,9\ см.\]

\[2)\ a = 18\ см,\ b = 25\ см,\ \alpha = 36{^\circ}:\]

\[\sin\beta = \frac{b\sin\alpha}{a} = \frac{25 \bullet \sin{36{^\circ}}}{18};\]

\[\sin\beta \approx 0,8164.\ \ \ \]

\[\beta_{1} \approx 55{^\circ};\ \ \ \beta_{2} \approx 125{^\circ}.\]

\[\gamma_{1} = 180{^\circ} - 55{^\circ} - 36{^\circ} = 89{^\circ};\]

\[c_{1} = \frac{a\sin\gamma}{\sin\alpha} =\]

\[= \frac{18 \bullet \sin{89{^\circ}}}{\sin{36{^\circ}}} \approx 30,6\ см.\]

\[\gamma_{2} = 180{^\circ} - 125{^\circ} - 36{^\circ} = 19{^\circ};\]

\[c_{2} = \frac{a\sin\gamma}{\sin\alpha} =\]

\[= \frac{18 \bullet \sin{19{^\circ}}}{\sin{36{^\circ}}} \approx 10,0\ см.\]