\(1.003\ sen\ 45=1.62\ sen\ \theta_1=\left(1.47\right)sen\theta_2\)
\(\tan\theta_1=\frac{x1}{2}=2\tan\theta_1\)
\(\tan\theta_2=\frac{x2}{3}=3\tan\theta_2\)
\(\theta1=sen^{-1}\left[\left(\frac{1.003}{1.62}\right)sen\ 45\ grados\right]\)=25.96
\(\theta2=sen^{-1}\left[\left(\frac{1.003}{1.47}\right)sen\ 45\ grados\right]\) =28-84
D=x1+x2
\(D=2\tan\theta_1+3\tan\theta_2\)
D=2.62 cm