\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[\angle BAC = 42{^\circ};\]
\[\angle ACB = 64{^\circ};\]
\[AC = 20\ м.\]
\[Найти:\]
\[\text{AB.}\]
\[Решение:\]
\[\angle B = 180{^\circ} - \angle A - \angle C = 74{^\circ};\]
\[\frac{\text{AC}}{\sin{\angle B}} = \frac{\text{AB}}{\sin{\angle C}}\text{\ \ }\]
\[AB = \frac{AC \bullet \sin{\angle C}}{\sin{\angle B}}\]
\[AB = 20 \bullet \frac{\sin{64{^\circ}}}{\sin{74{^\circ}}} \approx\]
\[\approx 20 \bullet \frac{0,898}{0,96} \approx 18,7\ м.\]
\[Ответ:\ \approx 18,7\ м.\]