Question 24·Hard·Right Triangles and Trigonometry

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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

For right triangles where an altitude is drawn from the right angle to the hypotenuse, immediately note that this splits the hypotenuse into two segments and creates two smaller right triangles similar to the original. Use similarity (or the memorized facts $DE^2 = DF\cdot DG$ and $EF^2 = DF\cdot GF$ ) to express each leg in terms of the hypotenuse and its adjacent segment, then form the tangent as opposite/adjacent. To save time and avoid messy square roots, work with squared lengths first (like $EF^2/DE^2$ ) and only take a square root at the end, and always double-check which leg is opposite and which is adjacent to the given angle before forming the tangent ratio.

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Focus on the hypotenuse and its segments

Draw (or imagine) triangle $DEF$ with a right angle at $E$ , and an altitude from $E$ down to the hypotenuse at $G$ . How long is the entire hypotenuse $DF$ in terms of $DG$ and $GF$ ?

Use the similar triangles created by the altitude

The altitude from the right angle to the hypotenuse creates two smaller right triangles that are each similar to the original triangle. Think about writing a proportion involving $DE$ , $DF$ , and $DG$ , and another involving $EF$ , $DF$ , and $GF$ .

Connect leg lengths to tangent

For angle $D$ in the original triangle, which leg is opposite and which is adjacent? Express $\tan(D)$ as a ratio of these legs, and consider how your equations for $DE^2$ and $EF^2$ can help you find this ratio without directly taking square roots at first.

Desmos Guide

Compute the ratio of hypotenuse segments

In Desmos, type 54/6 and note the value. This corresponds to the ratio $GF/DG$ that appears when you form $\dfrac{EF^2}{DE^2}$ .

Take the square root to get $\tan(D)$

In Desmos, type sqrt(54/6) (or sqrt(ans) if your calculator supports using the previous answer). The displayed value is the value of $\tan(D)$ .

Step-by-step Explanation

Understand the setup and find the hypotenuse

$\angle E$ is the right angle, so $\overline{DF}$ is the hypotenuse of $\triangle DEF$ .
The altitude from $E$ meets $\overline{DF}$ at $G$ , splitting the hypotenuse into two parts: $DG = 6$ and $GF = 54$ .
So the whole hypotenuse is

DF = DG + GF = 6 + 54 = 60.

Relate leg DE to DF and DG using similar triangles

Dropping an altitude from the right angle in a right triangle creates two smaller right triangles that are similar to the original triangle.

$\triangle DEF$ and $\triangle DEG$ are both right triangles and share angle $D$ , so they are similar.
From similarity, the ratio of a leg to the hypotenuse in the big triangle equals the ratio of the corresponding leg to the hypotenuse in the smaller triangle:

\frac{DE}{DF} = \frac{DG}{DE}.

Cross-multiply to get a relationship involving $DE^2$ :

DE^2 = DF \cdot DG = 60 \cdot 6.

Relate leg EF to DF and GF using similar triangles

Now use the other small triangle.

$\triangle DEF$ and $\triangle EGF$ are also similar (both are right triangles and share angle $F$ ).
So we have a similar proportion for $EF$ :

\frac{EF}{DF} = \frac{GF}{EF}.

Cross-multiply to get a relationship involving $EF^2$ :

EF^2 = DF \cdot GF = 60 \cdot 54.

Use the leg ratios to find $\tan(D)$

By definition in a right triangle,

For angle $D$ , the opposite side is $EF$ .
The adjacent side is $DE$ .

\tan(D) = \frac{\text{opposite}}{\text{adjacent}} = \frac{EF}{DE}.

To use the equations for $DE^2$ and $EF^2$ , square the tangent:

\tan^2(D) = \left(\frac{EF}{DE}\right)^2 = \frac{EF^2}{DE^2}.

Substitute the expressions we found:

\tan^2(D) = \frac{60 \cdot 54}{60 \cdot 6} = \frac{54}{6} = 9.

Since angle $D$ is acute, $\tan(D)$ is positive, so

\tan(D) = \sqrt{9} = 3.

The correct answer is $3$ .

Question 24·Hard·Right Triangles and Trigonometry

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Step-by-step Explanation

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Step-by-step Explanation