Во-первых, ознакомься с рисунком(Рис.1 - вложение)
1)
AC\cdot tgA=CB => AC= \frac{CB}{tgA}" alt="tgA= \frac{CB}{AC} => AC\cdot tgA=CB => AC= \frac{CB}{tgA}" align="absmiddle" class="latex-formula">
2)
AC= \frac{20\sqrt{10}}{10} => AC=2\sqrt{10}" alt="AC= \frac{3} {3 \sqrt {10}} \cdot 20 =>AC= \frac{20\sqrt{10}}{10} => AC=2\sqrt{10}" align="absmiddle" class="latex-formula">
3) По теореме Пифагора
![AC=\sqrt{(2\sqrt{10})^2 + 3^2 } = \sqrt{49} = 7 AC=\sqrt{(2\sqrt{10})^2 + 3^2 } = \sqrt{49} = 7](https://tex.z-dn.net/?f=AC%3D%5Csqrt%7B%282%5Csqrt%7B10%7D%29%5E2+%2B+3%5E2+%7D+%3D+%5Csqrt%7B49%7D+%3D+7)
Ответ: АС = 7 единиц