SAT Right Triangles & Trigonometry Question 21 | Free SAT Prep

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Question 21·Hard·Right Triangles and Trigonometry

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In right triangle $ABC$ shown, $\angle C$ is a right angle. Point $D$ lies on $\overline{AB}$ , and $\overline{CD}\perp \overline{AB}$ . If $AD=4$ and $DB=9$ , which choice is equal to

\frac{\cos(\angle ACD)}{\cos(\angle BCD)}?

Translate each cosine into a side ratio in the appropriate right triangle, and simplify the overall expression before plugging in any numbers. When an altitude is drawn to the hypotenuse of a right triangle, look for similar triangles; these give relationships like $AC^2=AD\cdot AB$ and $BC^2=DB\cdot AB$ , which make ratios such as $\frac{BC}{AC}$ depend only on $\frac{DB}{AD}$ .

Hints

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Use the altitude to identify two right triangles

Because $CD\perp AB$ , focus on right triangles $ACD$ and $BCD$ when writing the cosine of each angle at $C$ .

Look for cancellation

When you write $\cos(\angle ACD)$ and $\cos(\angle BCD)$ as fractions, both expressions involve $CD$ . See what happens to $CD$ in the ratio.

Relate $AC$ and $BC$ to $AD$ and $DB$

The altitude from the right angle to the hypotenuse creates similar triangles. Use a proportion from similarity to connect $AC^2$ with $AD$ and $BC^2$ with $DB$ .

Desmos Guide

Enter the segment lengths as variables

In Desmos, define the segments by typing:

a=4

b=9

Compute the squared ratio from similarity

Type the expression for $\frac{DB}{AD}$ :

b/a

Take the square root to get the ratio

Type:

sqrt(b/a)

Then match the resulting value to one of the answer choices.

Step-by-step Explanation

Rewrite each cosine using the right triangles

Because $CD\perp AB$ , triangles $ACD$ and $BCD$ are right triangles with the right angle at $D$ .

For $\angle ACD$ (angle at $C$ in triangle $ACD$ ),

\cos(\angle ACD)=\frac{CD}{AC}.

For $\angle BCD$ (angle at $C$ in triangle $BCD$ ),

\cos(\angle BCD)=\frac{CD}{BC}.

Simplify the ratio

\frac{\cos(\angle ACD)}{\cos(\angle BCD)} = \frac{\frac{CD}{AC}}{\frac{CD}{BC}} = \frac{BC}{AC}.

So the problem becomes finding $\frac{BC}{AC}$ .

Use similarity from the altitude to the hypotenuse

In a right triangle, drawing the altitude from the right angle to the hypotenuse creates two smaller right triangles that are each similar to the original triangle. Here,

$\triangle ACD \sim \triangle ABC$ and $\triangle BCD \sim \triangle ABC$ .

From $\triangle ACD \sim \triangle ABC$ :

\frac{AD}{AC}=\frac{AC}{AB} \quad\Rightarrow\quad AC^2=AD\cdot AB.

From $\triangle BCD \sim \triangle ABC$ :

\frac{DB}{BC}=\frac{BC}{AB} \quad\Rightarrow\quad BC^2=DB\cdot AB.

Form the needed ratio and substitute

Divide the two equations:

\frac{BC^2}{AC^2}= \frac{DB\cdot AB}{AD\cdot AB}= \frac{DB}{AD}.

\left(\frac{BC}{AC}\right)^2=\frac{9}{4} \quad\Rightarrow\quad \frac{BC}{AC}=\sqrt{\frac{9}{4}}.

Compute and choose the matching option

\sqrt{\frac{9}{4}}=\frac{3}{2}.

Therefore,

\frac{\cos(\angle ACD)}{\cos(\angle BCD)}=\frac{3}{2}.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation