Question 424·Hard·Math

The square at $D$ makes $AC$ the hypotenuse, the side across from the right angle in triangle $ACD$. From $\angle ACD$, $AD$ is opposite. Since sine is opposite divided by hypotenuse, the given ratio means $AD$ is the $5$ part:

The numbers $5,12,13$ form a Pythagorean triple: $5^2+12^2=13^2$, so they can be the sides of a right triangle. Here $AD$ matches $5$ and $AC$ matches $13$, so $CD$ matches $12$:

The $12$ part is $CD=60$, so multiply the $5$ part by $\frac{60}{12}$. Type $\frac{5}{12}\cdot60$ in Desmos; it prints $25$. So $AD=25$.

$CD$ is an altitude: it runs from the right angle at $C$ perpendicular to $AB$. It creates two similar right triangles, whose matching sides give the altitude squared equals the product of the two pieces of the hypotenuse:

Substitute the lengths you know:

Let $x_1$ represent $DB$. Type $60^2\sim25x_1$ in Desmos. The $\sim$ tells Desmos to find the value of $x_1$ that fits the equation. Under PARAMETERS, it shows $x_1=144$, so $DB=144$.

Point $D$ lies on $AB$, so add the two pieces along that segment:

Type $x_1+25$ in Desmos; it prints $169$. The length of $\overline{AB}$ is $169$. Choice D.

Triangles $ACD$ and $ABC$ share angle $A$, and both are right triangles. Their remaining acute angles must match, so $\angle ACD=\angle ABC$. You can use the given sine ratio at angle $B$ of the large triangle.

In the $5:12:13$ ratio, $AC$ is the $13$ part and $CD=60$ is the $12$ part. Type $\frac{13}{12}\cdot60$ in Desmos; it prints $65$. So $AC=65$, the side opposite angle $B$ in the large triangle.

In triangle $ABC$, sine at $B$ is opposite over hypotenuse, so the denominator is $AB$:

Substitute $AC=65$:

Let $x_1$ represent $AB$ and type $\frac{65}{x_1}\sim\frac{5}{13}$ in Desmos. Under PARAMETERS, it shows $x_1=169$, so $AB=169$. Choice D.