Right triangles and trigonometry SAT question

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 $\text{tan}(A)=\frac{50}{7}$, then $\text{tan}(D)=\frac{50}{7}$. 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, $\text{tan}(E)=\frac{1}{\text{tan}(D)}$. Substituting $\frac{50}{7}$ for $\text{tan}(D)$ in this equation yields $\text{tan}(E)=\frac{1}{\frac{50}{7}}$, or $\text{tan}(E)=\frac{7}{50}$. Thus, if $\text{tan}(A)=\frac{50}{7}$, the value of $\text{tan}(E)$ is $\frac{7}{50}$. Note that 7/50 and .14 are examples of ways to enter a correct answer.