[tex]\it \dfrac{CD}{AD}=\dfrac{3}{4} \Rightarrow \dfrac{CD^2}{AD^2}=\dfrac{9}{16} \stackrel{T.h}{\Rightarrow} \dfrac{CD^2}{BD\cdot CD}=\dfrac{9}{16} \Rightarrow\dfrac{CD}{BD}=\dfrac{9}{16} \Rightarrow
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\dfrac{CD+BD}{BD}=\dfrac{9+16}{16} \Rightarrow \dfrac{BC}{BD}=\dfrac{25}{16} \Rightarrow BD = \dfrac{16BC}{25} \ \ \ \ (1)[/tex]
[tex]\it T.\ catetei\ \Rightarrow AB^2=BC\cdot BD \ \ \ \ (2) \\\;\\ (1),\ (2) \Rightarrow AB^2 =BC \cdot \dfrac{16BC}{25} \Rightarrow AB^2 = BC^2\cdot \dfrac{4^2}{5^2} \Rightarrow[/tex]
[tex]\it \Rightarrow AB = BC\cdot \dfrac{4}{5} \Rightarrow 40= BC\cdot \dfrac{4}{5} \Rightarrow BC = 40\cdot \dfrac{5}{4} =10\cdot5 =50 cm[/tex]
Cu teorema lui Pitagora se determină AC = 30 cm.
[tex]\it \mathcal{A} = \dfrac{AB\cdot AC}{2} = \dfrac{40\cdot30}{2} =20\cdot30 = 600 cm^2[/tex]
sinB = cateta opusă/ipotenuză
sinB = AC/BC = 30/50 = 3/5 = 0,6
