Question 51·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

Because $\angle B$ is right, $AC$ is the hypotenuse, the side across from that angle. An altitude meets its base at a right angle, so $BD$ makes both $\triangle ABD$ and $\triangle BCD$ right triangles at $D$.

From $\angle A$, $BD$ is opposite and $AB$ is the hypotenuse. Sine is opposite over hypotenuse, so $\sin A=\frac{BD}{AB}=\frac{12}{13}$.

The Pythagorean theorem says the squares of the two legs add to the hypotenuse's square. Since $5^2+12^2=13^2$, the remaining leg has the $5$ part. So $AD:BD=5:12$.

Use $BD=120$ for the $12$ part. Type $120\cdot\frac{5}{12}$ in Desmos; it displays $50$. So $AD=50$, the piece next to $A$, not the piece the question asks for.

The two smaller triangles are similar: both are right at $D$, and $\angle ABD$ equals $\angle BCD$ because each complements $\angle A$. Their matching sides give the altitude squared equals the product of the two hypotenuse pieces: $BD^2=AD\cdot DC$.

Divide by $AD$ to isolate the requested piece: $DC=\frac{BD^2}{AD}$.

Substitute $BD=120$ and $AD=50$: type $\frac{120^2}{50}$ in Desmos. It displays $288$, so the length of $\overline{DC}$ is $288$. Choice C.