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 $\cos x$ ?

Note: Figure not drawn to scale.

An altitude to the hypotenuse splits a right triangle into two smaller right triangles. To find a cosine ratio, first identify the adjacent side in the triangle containing the angle. Then use similarity to find the pieces of the hypotenuse; an inequality tells you which piece is longer. Swapping adjacent and opposite gives sine instead of cosine.

Hints

Hint 1
Cosine is adjacent over hypotenuse in a right triangle. In the small triangle containing $x$ , $D$ is the right angle. Which side touches $x$ without being the hypotenuse?
Hint 2
The two small triangles share the leg $CD$ . By the Pythagorean theorem, the larger of their hypotenuses, $AC$ and $BC$ , goes with the larger piece of $AB$ . Which piece exceeds half of $30$ ?
Hint 3
The altitude from $C$ to the hypotenuse gives $CD^2=AD\cdot DB$ . Write one piece as $30$ minus the other, then keep the solution that satisfies $AC>BC$ .

Step-by-step

Find the hypotenuse next to the angle
Step 1
Identify the cosine ratio
Since $CD$ is perpendicular to $AB$ , $\triangle ACD$ is right at $D$ . For $x=\angle ACD$ , $CD=12$ is the adjacent leg, the leg touching $x$ , and $AC$ is the hypotenuse, across from the right angle. So cosine is adjacent over hypotenuse: $\cos x=\frac{12}{AC}$ .
Step 2
Determine which piece is longer
Both small triangles have $CD=12$ as a leg. The Pythagorean theorem says the legs' squares add to the hypotenuse's square: $AC^2=AD^2+12^2$ , $BC^2=DB^2+12^2$ . Since $AC>BC$ , the first triangle must have the longer other leg: $AD>DB$ . Also, $AD+DB=AB=30$ , so $AD>15$ . The longer leg of the original triangle touches the longer piece of the hypotenuse.
Step 3
Relate the altitude to the two pieces
The two small right triangles are similar: they have the same angle sizes, even though they face different ways. Match their corresponding sides: $\frac{CD}{AD}=\frac{DB}{CD}$ . Cross-multiply to relate the altitude to the two pieces: $CD^2=AD\cdot DB$ .
Step 4
Write an equation for the longer piece
Because the pieces total $30$ , $DB=30-AD$ . Substitute that and $CD=12$ into the altitude relationship: $12^2=AD(30-AD)$ .
Step 5
Solve for the piece next to the longer leg
Type $12^2\sim x_1(30-x_1)\{x_1>15\}$ , where $x_1$ stands for $AD$ . The $\sim$ tells Desmos to find a value that fits the equation; the restriction keeps the longer piece, as required by $AC>BC$ . Under PARAMETERS, Desmos shows $x_1=24$ , so $AD=24$ .
Step 6
Find the hypotenuse of the small triangle
Use $AD=24$ in the Pythagorean relationship for $\triangle ACD$ . Type $\sqrt{x_1^2+12^2}$ ; Desmos shows about $26.833$ . Keep the exact length for the radical choices: $AC=\sqrt{24^2+12^2}$ $AC=\sqrt{720}$ $AC=12\sqrt5$ .
Step 7
Evaluate cosine and match its exact form
Type $\frac{12}{\sqrt{x_1^2+12^2}}$ and $\frac{\sqrt5}{5}$ ; Desmos shows about $0.4472$ for each. To see the exact match, divide the cosine ratio by $12$ : $\cos x=\frac{12}{12\sqrt5}=\frac{1}{\sqrt5}$ . Multiply its numerator and denominator by $\sqrt5$ : $\cos x=\frac{\sqrt5}{5}$ . Choice D.

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

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