Question 540·Medium·Math

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In right triangle $DEF$ , angle $E$ is the right angle and $\tan(D)=\dfrac{2}{5}$ . What is $\cos(F)$ ?

A tangent ratio gives the two legs of a right triangle up to a shared scale. If the question asks for cosine of the other acute angle, switch which leg is adjacent, then find the hypotenuse with the Pythagorean theorem. Desmos can handle the squares; don't mistake the given tangent for the requested cosine.

Hints

Hint 1
A tangent ratio compares the leg opposite an angle with the leg adjacent to it. From angle $D$ , which sides make the given ratio? You can use $2$ and $5$ as representative lengths because scaling both legs won't change the answer.
Hint 2
A cosine ratio compares the adjacent leg with the hypotenuse. Move from angle $D$ to angle $F$ : the legs switch roles, but the hypotenuse stays across from the right angle.
Hint 3
The Pythagorean theorem says the squares of the two legs add to the square of the hypotenuse. Use your representative leg lengths to find the hypotenuse, which cosine needs.

Step-by-step

Use the leg ratio and find the hypotenuse
Step 1
Turn tangent into representative leg lengths
Tangent means opposite over adjacent. From angle $D$ , $EF$ is opposite and $DE$ is adjacent, so $\frac{EF}{DE}=\frac{2}{5}$ . This fixes a ratio, not actual lengths. Choose $EF=2$ and $DE=5$ ; scaling both legs by the same amount won't change cosine.
Step 2
Identify the sides cosine of $F$ uses
Because $E$ is the right angle, $DF$ is the hypotenuse, the side across from it. From $F$ , $EF$ is adjacent, meaning it's the leg touching that angle. Changing acute angles swaps the legs' roles, not the hypotenuse. Cosine is adjacent over hypotenuse, so $\cos(F)=\frac{EF}{DF}=\frac{2}{DF}$ .
Step 3
Find the hypotenuse
The Pythagorean theorem says the squares of the legs add to the square of the hypotenuse: $DF^2=2^2+5^2$ . Type $2^2+5^2$ in Desmos; it shows $29$ . Because a side length is positive, $DF=\sqrt{29}$ .
Step 4
Write cosine in the answer's form
Substitute the hypotenuse:
$\cos(F)=\frac{2}{\sqrt{29}}.$
To rationalize, or remove the root from the bottom, multiply by $1$ written as $\frac{\sqrt{29}}{\sqrt{29}}$ :
$\frac{2}{\sqrt{29}}\cdot\frac{\sqrt{29}}{\sqrt{29}}=\frac{2\sqrt{29}}{29}.$
Type both fractions in Desmos; each shows about $0.3714$ . So the cosine of $F$ is $\frac{2\sqrt{29}}{29}$ . Choice D.

Question 540·Medium·Math

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