Question 139·Medium·Geometry and Trigonometry

Remember $\tan(\theta) = \dfrac{\text{opposite}}{\text{adjacent}}$. If $\tan(D) = \dfrac{2}{5}$, which sides of the triangle could you set equal to 2 and 5?

Once you know the two legs, use the Pythagorean theorem to find the hypotenuse. Then, for angle $F$, identify which side is adjacent and which is the hypotenuse to set up $\cos(F)$.

Check whether you’re taking cosine with respect to angle $D$ or angle $F$. The adjacent side changes depending on which angle you’re looking at.

In a Desmos expression line, type D = arctan(2/5). This defines an angle $D$ whose tangent is $2/5$.

Because $D$ and $F$ are acute angles in a right triangle, $F = \frac{\pi}{2} - D$ in radians. In Desmos, type F = pi/2 - D to define $F$.

In a new line, type cos(F). Note the decimal value Desmos gives you; this is the numerical value of $\cos(F)$.

In separate lines, enter each expression from the choices: 2*sqrt(29)/29, 5*sqrt(29)/29, 2/5, and 5/2. Compare their decimal values to the value of cos(F); the matching expression corresponds to the correct choice.

In right triangle $DEF$, angle $E$ is $90^\circ$, so $DF$ is the hypotenuse and $DE$, $EF$ are the legs.

We are told $\tan(D)=\dfrac{2}{5}$. By definition,

For angle $D$:

• The side opposite $D$ is $EF$.
• The side adjacent to $D$ (but not the hypotenuse) is $DE$.

So we can take

• $EF = 2$ and
• $DE = 5$

(or any multiples of these; the ratio is what matters).

Now find $DF$, the hypotenuse, using the Pythagorean theorem:

So

Cosine of an angle is

For angle $F$:

• The hypotenuse is still $DF = \sqrt{29}$.
• The leg adjacent to $F$ (that is not the hypotenuse) is $EF = 2$.

So

Rationalize the denominator of $\dfrac{2}{\sqrt{29}}$:

This matches answer choice D, so the correct answer is $\dfrac{2\sqrt{29}}{29}$.