Question

Решите номер 12.Есть вложение.

Answer 1

$\cos A=\frac{AC}{AB}=\frac{AC}{\sqrt{AC^2+BC^2}} \\\ \sqrt{AC^2+BC^2}=\frac{AC}{\cos A} \\\ AC^2+BC^2=\frac{AC^2}{\cos^2 A} \\\ \frac{AC^2}{\cos^2A}-AC^2=BC^2 \\\ AC^2(\frac{1}{\cos^2A}-1)=BC^2 \\\ AC^2=\cfrac{BC^2}{\frac{1}{\cos^2A}-1} \\\ AC=\sqrt{\cfrac{BC^2}{\frac{1}{\cos^2A}-1}} \\\ AC=\sqrt{\cfrac{6^2}{\frac{1}{(\frac{3\sqrt{13}}{13})^2}-1}}= \sqrt{\cfrac{36}{\frac{1}{\frac{9}{13}}-1}}= \sqrt{\cfrac{36}{\frac{13}{9}-1}}= \sqrt{\cfrac{36}{\frac{4}{9}}}= \sqrt{81}=9$
Ответ: 9