[tex] \it AO = \dfrac{AC}{2} = \dfrac{8}{2} =4\ cm\\ \\ \\AC \perp BD \Rightarrow AO \perp BD \Rightarrow AO - \^{i}n\breve{a}l\c{\it t}ime \ pentru \ \Delta ABD [/tex]
[tex] \it AO= \dfrac{AC}{2} =\dfrac{8}{2} = 4\ cm\\ \\ \\\mathcal{A}_{ABD} = \dfrac{BD\cdot AO}{2} = \dfrac{6\cdot4}{2} = 12\ cm^2\\ \\ \\ Dar,\ \ \mathcal{A}_{ABD} = \dfrac{AB\cdot AD \cdot sinA}{2} \Rightarrow 12 = \dfrac{AB\cdot AD \cdot sinA}{2} \Rightarrow 24 = AB\cdot AD\cdot sinA \\ \\ \\ \Rightarrow sinA = \dfrac{24}{AB\cdot AD} \ \ \ \ (1) [/tex]
Deoarece diagonalele se înjumătățesc, triunghiul dreptunghic AOB
are catetele AO = 4cm, BO = 3 cm ⇒ AB = 5 cm (cu Th. Pitagora).
Laturile rombului au lungimile egale, deci AD = AB = 5 cm (2)
[tex] \it (1), (2) \Rightarrow sinA = \dfrac{24}{5\cdot 5} =\dfrac{24}{25} [/tex]
