Question 496·Hard·Math

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In right triangle $DEF$ , $\angle E$ is the right angle. The altitude from $E$ to hypotenuse $\overline{DF}$ meets $\overline{DF}$ at point $G$ so that $DG = 6$ and $GF = 54$ . What is $\tan(D)$ ?

Your answer

An altitude from the right angle to the hypotenuse creates two similar triangles. Their matching sides let you find the altitude from the two pieces of the hypotenuse. Then use the small right triangle containing the requested angle: tangent is its opposite leg divided by its adjacent leg. Finding the altitude alone stops one move too soon.

Hints

Hint 1
The altitude meets the hypotenuse at a right angle, so the small triangle $DEG$ is a right triangle containing $\angle D$ . Which leg is opposite $D$ , and which leg touches it?
Hint 2
The two small triangles are similar: both have a right angle at $G$ , and their other angles match those in the large triangle. Match sides by their angles, not by where the sides appear.
Hint 3
In similar triangles, matching side ratios are equal. Pair $DG$ with $EG$ , and pair $EG$ with $GF$ . The resulting proportion finds the altitude, which you'll still need to divide by $DG$ .

Step-by-step

Use the two smaller triangles
Step 1
Choose the triangle for tangent
Because $G$ lies on $DF$ , triangle $DEG$ contains the same $\angle D$ as the large triangle. In this smaller right triangle, $EG$ is opposite $D$ , and $DG$ is the adjacent leg that touches $D$ . Tangent means opposite over adjacent, so:
$\tan(D)=\frac{EG}{DG}=\frac{EG}{6}.$
Step 2
Match the smaller triangles
The altitude makes both small triangles right triangles at $G$ . In triangle $DEG$ , $\angle DEG$ complements $\angle D$ ; in the large triangle, $\angle EFG$ does too. So those angles are equal. By AA similarity, meaning two pairs of equal angles, $\triangle DGE\sim\triangle EGF$ . Their matching sides are $DG\leftrightarrow EG$ and $EG\leftrightarrow GF$ .
Step 3
Write the matching-side proportion
Matching sides of similar triangles have equal ratios, so write:
$\frac{EG}{DG}=\frac{GF}{EG}.$
Put in the given pieces, $DG=6$ and $GF=54$ :
$\frac{EG}{6}=\frac{54}{EG}.$
Step 4
Relate the altitude to the pieces
Multiply both sides by $6EG$ to remove the denominators:
$EG^2=6(54).$
The altitude squared is the product of the two hypotenuse pieces. This finds $EG$ , not the tangent yet.
Step 5
Find the positive altitude
Since $EG$ is a length, keep the positive solution. Type $x_1^2\sim6(54)\{x_1>0\}$ in Desmos, using $x_1$ for $EG$ ; $\sim$ asks Desmos to find the value that fits the equation. Under PARAMETERS, it shows $x_1=18$ , so $EG=18$ .
Step 6
Finish the tangent ratio
Return to triangle $DEG$ . Its opposite leg is $EG=18$ , and its adjacent leg is $DG=6$ . Divide, rather than stopping at the altitude: type $x_1/6$ in Desmos, using $x_1$ for $EG$ ; Desmos shows $3$ .
$\tan(D)=\frac{EG}{DG}=\frac{18}{6}=3.$
The tangent of $\angle D$ is $3$ . Grid in 3.

Question 496·Hard·Math

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