Question 150·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

0 / 200 Questions completed

In right triangle $ABC$ , angle $C$ is $90^\circ$ . The altitude from $C$ to hypotenuse $\overline{AB}$ meets $\overline{AB}$ at point $D$ , and $\overline{CD}\perp \overline{AB}$ . If $CD=15$ and $\tan(A)=\frac{3}{4}$ , which choice is the length of $AD$ ?

An altitude to a right triangle’s hypotenuse creates two smaller right triangles. If the question asks for one piece of the hypotenuse, look for the small triangle containing that piece and the given length. Use tangent, opposite over adjacent, to write a ratio, then solve it in Desmos. The trap is finding the whole hypotenuse instead of the requested piece.

Hints

Hint 1
The altitude meets the hypotenuse at a right angle. Since $AD$ lies on $AB$ , triangle $ACD$ is a right triangle that shares angle $A$ with the original triangle. Look at this smaller triangle.
Hint 2
In triangle $ACD$ , $CD$ is opposite angle $A$ , and $AD$ is adjacent to it. Tangent compares those two legs: opposite divided by adjacent. How can you write the given tangent using $CD$ and $AD$ ?
Hint 3
To solve for a length in Desmos, replace $AD$ with $x_1$ and change $=$ to $\sim$ . Desmos finds the value of $x_1$ that makes the ratio true. Keep track of which segment $x_1$ represents.

Step-by-step

Use tangent in the smaller triangle
Step 1
Choose the right triangle containing $AD$
Because the altitude $CD$ is perpendicular to $AB$ at $D$ , triangle $ACD$ is right at $D$ . Since $AD$ lies on $AB$ , this triangle also has the original angle $A$ . Use the smaller triangle containing both the given $CD$ and the requested $AD$ .
Step 2
Write the tangent ratio
In triangle $ACD$ , $CD=15$ is across from angle $A$ , while $AD$ is the leg beside it. Tangent means opposite divided by adjacent, so:
$\tan(A)=\frac{CD}{AD}$
Substitute the given ratio and length:
$\frac{3}{4}=\frac{15}{AD}.$
Step 3
Solve for the requested segment
Type $\frac{15}{x_1}\sim\frac{3}{4}$ . Here $x_1$ stands for $AD$ , and $\sim$ tells Desmos to find the value that makes the sides equal. Under PARAMETERS, Desmos shows $x_1=20$ . So $AD$ has length $20$ . Choice A.

Question 150·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this