Triangle A B C is similar to triangle D E F , where angle A corresponds

The correct answer is $\frac{7}{50}$. It's given that triangle $ABC$ is similar to triangle $DEF$, where angle $A$ corresponds to angle $D$ and angle $C$ corresponds to angle $F$. In similar triangles, the tangents of corresponding angles are equal. Since angle $A$ and angle $D$ are corresponding angles, if $\tan \left(A\right)=\frac{50}{7}$, then $\tan \left(D\right)=\frac{50}{7}$. It's also given that angles $C$ and $F$ are right angles. It follows that triangle $DEF$ is a right triangle with acute angles $D$ and $E$. The tangent of one acute angle in a right triangle is the inverse of the tangent of the other acute angle in the triangle. Therefore, $\tan \left(E\right)=\frac{1}{\tan \left(D\right)}$. Substituting $\frac{50}{7}$ for $\tan \left(D\right)$ in this equation yields $\tan \left(E\right)=\frac{1}{{\displaystyle \frac{50}{7}}}$, or $\tan \left(E\right)=\frac{7}{50}$. Thus, if $\tan \left(A\right)=\frac{50}{7}$, the value of $\tan \left(E\right)$ is $\frac{7}{50}$. Note that 7/50 and .14 are examples of ways to enter a correct answer.