Na figura acima, ABC e AED são triângulos retângulos. Se m(AC) = l , m(BÂC) = α , \(m\left(A\hat DE\right)\)\(=\beta\) e \(m\left(A\hat BC\right)=m\left(D\hat AE\right)=90^0\), então m(BD) é
l ⋅ cos α
l ⋅ sen2α
l ⋅ cos α ⋅ sen β
\(\frac{l.cos^2\alpha}{sen\beta}\)
\(\frac{l.sen^2\alpha}{cos\beta}\)