Question #2be01bd9 - SAT Suite Question Bank

ID: 2be01bd9

Modded SAT Question Bank by Abdullah Mallik

Triangle $A B C$ is similar to triangle $D E F$ , where angle $A$ corresponds to angle $D$ and angle $C$ corresponds to angle $F$ . Angles $C$ and $F$ are right angles. If $\tan (A) = \frac{50}{7}$ , what is the value of $\tan (E)$ ?

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Correct Answer: .14, 7/50

Rationale

The correct answer is $\frac{7}{50}$ . It's given that triangle $A B C$ is similar to triangle $D E F$ , 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 $\tan (A) = \frac{50}{7}$ , then $\tan (D) = \frac{50}{7}$ . It's also given that angles $C$ and $F$ are right angles. It follows that triangle $D E F$ 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, $\tan (E) = \frac{1}{\tan (D)}$ . Substituting $StartFraction 50 Over 7 EndFraction$ for $\tan (D)$ in this equation yields $\tan (E) = \frac{1}{\frac{50}{7}}$ , or $\tan (E) = \frac{7}{50}$ . Thus, if $\tan (A) = \frac{50}{7}$ , the value of $\tan (E)$ is $\frac{7}{50}$ . Note that 7/50 and .14 are examples of ways to enter a correct answer.

Question Difficulty: Hard