Folosim formula fundamentală a trigonometriei:
[tex]\it sin^2x+cos^2=1 \Rightarrow sin^2x=1-cos^2x[/tex]
Pentru problema dată, vom avea:
[tex]\it sin^2(ABC) = 1-cos^2(ABC) = 1-\Big(\dfrac{4}{5}\Big)^2=1-\dfrac{16}{25} =\dfrac{25}{25}-\dfrac{16}{25} = \dfrac{^{4)}9}{\ 25} =\\ \\ \\ =\dfrac{36}{100}\ \ \ (1)\\ \\ \\ Dar,\ sin(ABC) =\dfrac{AC}{BC} \Rightarrow sin^2(ABC)= \dfrac{AC^2}{100}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow \dfrac{AC^2}{100}=\dfrac{36}{100} \Rightarrow AC^2=36=6^2 \Rightarrow AC = 6\ cm.[/tex]
