Use $\sin(A) = \dfrac{5}{13}$ to label the side opposite $A$ and the hypotenuse with convenient numbers. Which side should be $5$ and which should be $13$?

Once you know two sides of a right triangle (one of them the hypotenuse), use the Pythagorean theorem to find the third side. That third side will be both adjacent to $A$ and opposite $B$.

From angle $B$'s perspective, identify which side is opposite and which is adjacent (not the hypotenuse), then form the ratio opposite/adjacent to get $\tan(B)$.

In a new expression line, type: A = arcsin(5/13). Desmos will interpret this in radians and store that angle as A.

In the next line, type: B = pi/2 - A. This sets B to $90^\circ - A$ (since $\pi/2$ radians equals $90^\circ$).

In a new line, type: tan(B). Note the decimal value that Desmos gives for this expression.

Type each answer choice fraction (5/12, 12/5, 5/13, 13/12) into separate lines in Desmos and compare their decimal values to the value of tan(B). The matching fraction is the correct answer choice.

In right triangle $ABC$, angle $C$ is the right angle, so the other two angles are acute and must add to $90^\circ$.

That means

so $B$ is the complement of $A$ and $B = 90^\circ - A$.

By definition, for angle $A$ in a right triangle,

• $\sin(A) = \dfrac{\text{opposite to } A}{\text{hypotenuse}}$.

We are given $\sin(A) = \dfrac{5}{13}$, so we can take:

• side opposite angle $A$ to be $5$,
• the hypotenuse to be $13$.

Now the third side (adjacent to $A$ and opposite $B$) is still unknown.

Let $x$ be the length of the third side (adjacent to $A$ and opposite $B$). Since the triangle is right with hypotenuse $13$:

So the sides of the triangle form a $5$–$12$–$13$ right triangle.

For angle $B$, tangent is defined as

• $\tan(B) = \dfrac{\text{opposite to } B}{\text{adjacent to } B}$.

From angle $B$'s point of view:

• the side opposite $B$ is the side of length $12$,
• the side adjacent to $B$ (but not the hypotenuse) is the side of length $5$.

Therefore,

which matches answer choice B.