[tex] 30^2+40^2=50^2\\\text{Conform Reciprocei Teoremei lui Pitagora}\rightarrow [/tex][tex] \bigtriangleup DE F \text{ dreptunghic, cu }m(\sphericalangle E)=90^\circ\\\mathcal{A}_{\bigtriangleup DE F}=\frac{c_1\cdot c_2}{2}=\frac{30\cdot 40}{2}=600.\\\\\text{Distan\c ta de la un punct la o dreapta este}\\\text{ perpendiculara dus\u a din punct pe dreapta=\^in\u al\c timea}\\\text{Vom nota punctul cu A}\\\mathcal{A}_{\bigtriangleup DE F}=\frac{baza\cdot \^in\u al timea}{2}\\\frac{50\cdot EA}{2}=600\rightarrow EA=24\\\\EM \text{ median\u a}\rightarrow EM=\frac{DF}{2}=25 [/tex]
[tex] \text{In triunghiul dreptunghic MAE}\\\sin{\sphericalangle M}=\frac{cateta\: opus\u a}{ipotenuz\u a}=\frac{EA}{EM}=\frac{24}{25} [/tex]
