In right triangle , angle is a right angle. Point lies on such that . The length of is 30, the length of is 12, and . If , which choice is the value of ? Note: Figure not drawn to scale.

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Super-Hard SAT Math Question 138 | Free SAT Prep | Aniko

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Question 138·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In right triangle $ABC$ , angle $C$ is a right angle. Point $D$ lies on $\overline{AB}$ such that $\overline{CD}\perp \overline{AB}$ . The length of $\overline{AB}$ is 30, the length of $\overline{CD}$ is 12, and $AC>BC$ .

If $x=\angle ACD$ , which choice is the value of $\cos x$ ? © ‍Anіkο

Note: Figure not drawn to scale.

When an altitude is drawn to the hypotenuse of a right triangle, area is the quickest bridge between the hypotenuse/altitude information and the unknown legs: $\frac12(AC)(BC)=\frac12(AB)(CD)$ . Pair that with the Pythagorean theorem to get a solvable system. After you find the needed side length, compute the trig ratio in the correct right triangle (here, triangle $ACD$ ), not automatically in the original triangle. Ѕo⁠urce: аnіkо‍.aі

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Use two area formulas

Write the area of triangle $ABC$ once using the two legs $AC$ and $BC$ , and again using base $AB$ with height $CD$ . Сontеnt bу‌ ⁠Аniк‍о.а‌і

Create a system for the legs

You should get both $AC\cdot BC$ and $AC^2+BC^2$ . Those two facts together are enough to determine $AC$ and $BC$ .

Use the smaller right triangle for the cosine

Angle $x$ is in triangle $ACD$ . Identify the side adjacent to $x$ and the hypotenuse of triangle $ACD$ .

Desmos Guide

Model the legs with a variable

Let $x$ represent $AC$ (so $x>0$ ). Since $AB=30$ , use $BC=\sqrt{900-x^2}$ .

Use the area (product) condition

Enter the equation

x\cdot \sqrt{900-x^2}=360

Desmos will show one or two solutions for $x$ .

Choose the solution matching $AC>BC$

Click the solution points (or trace) to get the $x$ -values. Keep the value where $x>\sqrt{900-x^2}$ (because $AC>BC$ ). © аnіko.ai

Compute the cosine expression

In a new line, enter

\frac{12}{x}

using the chosen value of $x$ . The displayed value matches one of the answer choices.

Step-by-step Explanation

Relate the altitude to the area

The area of right triangle $ABC$ can be written two ways:

Using the legs $AC$ and $BC$ : Ѕоurс⁠e: ⁠аnіk‍o.‌аі

\text{Area}=\frac{1}{2}(AC)(BC)

Using hypotenuse $AB$ as the base and altitude $CD$ :

\text{Area}=\frac{1}{2}(AB)(CD)

Set them equal:

\frac{1}{2}(AC)(BC)=\frac{1}{2}(30)(12)

(AC)(BC)=360

Use the Pythagorean theorem

Because $\angle C$ is a right angle,

AC^2+BC^2=AB^2=30^2=900

Solve for the legs

Let $a=AC$ and $b=BC$ . Then

a^2+b^2=900

and

ab=360

Square the product: $a^2b^2=(360)^2=129600$ .

Now $a^2$ and $b^2$ are the two numbers whose sum is 900 and whose product is 129600. So $a^2$ and $b^2$ are solutions to

t^2-900t+129600=0

The discriminant is

900^2-4(129600)=810000-518400=291600

and $\sqrt{291600}=540$ , so

t=\frac{900\pm 540}{2}=720\text{ or }180

Thus the legs are $\sqrt{720}$ and $\sqrt{180}$ . Since $AC>BC$ , we take

AC=\sqrt{720}=12\sqrt{5}

Compute $\cos x$ in right triangle $ACD$

Triangle $ACD$ is a right triangle with right angle at $D$ (because $CD\perp AB$ and $AD$ lies on $AB$ ). For angle $x=\angle ACD$ , the adjacent side is $CD$ and the hypotenuse is $AC$ , so

\cos x=\frac{CD}{AC}=\frac{12}{12\sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}

Therefore, the correct choice is $\frac{\sqrt{5}}{5}$ .

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