[tex]\it AD=15cm \Rightarrow AB=BC=15 cm\\ \\ sinA=\dfrac{4}{5} \Rightarrow cosA=\sqrt{1-sin^2A}=\sqrt{1-\bigg(\dfrac{4}{5}\bigg)^2}=\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\\ \\ \\ \widehat B=180^o-\widehat A \Rightarrow cosB=-cosA=-\dfrac{3}{5}\\ \\ \\ Teorema \ lui\ Pitagora\ generalizat\breve a\ \hat\imath n\ \Delta ABC:\\ \\ AC^2=AB^2+BC^2-2AB\cdot BC\cdot cosB=15^2+15^2-2\cdot15\cdot15\cdot(-\dfrac{3}{5})=\\ \\ =225+225+2\cdot15\cdot3\cdot3=450+270=720[/tex]
[tex]\it AC=\sqrt{720}=\sqrt{144\cdot5}=12\sqrt5\ cm[/tex]
