Question 484·Hard·Math

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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)}?

An altitude to the hypotenuse splits a right triangle into smaller right triangles with the same shape. For a quotient of cosines, write adjacent over hypotenuse from each named angle and cancel any shared side. Then use the matching triangles to compare the sides left over. If you find a ratio of squared lengths, take a square root before choosing a ratio of lengths.

Hints

Hint 1
The perpendicular segment makes each smaller triangle a right triangle. Cosine means adjacent side over hypotenuse. From each angle at $C$ , which side is adjacent, and which side lies across from the right angle at $D$ ?
Hint 2
The two smaller triangles are similar to the large one: each shares an acute angle with it, and both have a right angle. Use matching sides to relate each leg of the large triangle to the hypotenuse piece beside it.
Hint 3
Dividing the two leg equations cancels the whole hypotenuse. That gives a ratio of squared legs, but the cosine quotient needs a ratio of leg lengths. What operation changes one into the other?

Step-by-step

Cancel the shared side, then use similarity
Step 1
Write each cosine as a side ratio
Because $CD$ is perpendicular to $AB$ , both small triangles have a right angle at $D$ . Cosine is adjacent leg over hypotenuse. For each angle at $C$ , $CD$ is adjacent, while $AC$ or $BC$ is the hypotenuse:
$\cos(\angle ACD)=\frac{CD}{AC},\qquad \cos(\angle BCD)=\frac{CD}{BC}.$
Step 2
Cancel the side both cosines use
Divide the fractions by multiplying by the reciprocal of the second one, then cancel $CD$ :
$\begin{aligned}\frac{\cos(\angle ACD)}{\cos(\angle BCD)}&=\frac{CD}{AC}\cdot\frac{BC}{CD}\\[1.4em]&=\frac{BC}{AC}.\end{aligned}$
So the target is a ratio of the two legs of the large triangle, not a ratio involving $CD$ .
Step 3
Match sides in the similar triangles
The large triangle and each small triangle share an acute angle and each have a right angle, so they're similar, or the same shape at different sizes. Match sides by their angles, not their position in the picture. For $\triangle ABC$ and $\triangle ACD$ , the hypotenuses are $AB$ and $AC$ , while $AC$ matches $AD$ . For $\triangle ABC$ and $\triangle CBD$ , $AB$ matches $BC$ , while $BC$ matches $DB$ . So:
$\frac{AB}{AC}=\frac{AC}{AD},\qquad \frac{AB}{BC}=\frac{BC}{DB}.$
Step 4
Turn the proportions into leg equations
Cross-multiply each proportion. Each leg's square uses the whole hypotenuse and the piece beside that leg:
$AC^2=AB\cdot AD,$ $BC^2=AB\cdot DB.$
Step 5
Cancel the whole hypotenuse
Divide the equation for $BC^2$ by the one for $AC^2$ :
$\frac{BC^2}{AC^2}=\frac{AB\cdot DB}{AB\cdot AD}.$
Cancel $AB$ , which both equations use:
$\frac{BC^2}{AC^2}=\frac{DB}{AD}.$
Step 6
Change squared lengths back to lengths
Both legs are positive lengths, so take the positive square root:
$\frac{BC}{AC}=\sqrt{\frac{DB}{AD}}.$
A ratio of squared legs needs a square root to become a ratio of lengths.
Step 7
Evaluate the ratio the question asks for
Use $DB=9$ and $AD=4$ . Type $\sqrt{\frac{9}{4}}$ in Desmos; it shows $1.5$ , and its fraction button shows $\frac{3}{2}$ . The cosine quotient equals $\frac{BC}{AC}=\frac{3}{2}$ . Choice A.

Question 484·Hard·Math

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