SAT Question 98e02472: Right Triangles And Trigonometry answer and explanation with Desmos solution

It's given that in right triangle $XYZ$, $\angle X$ and $\angle Z$ are acute angles. It follows that $\angle Y$ is a right angle. It's given that point $R$ lies on $XZ$ such that $YR$ is perpendicular to $XZ$. Since $\angle Y$ is a right angle and is comprised of $\angle XYR$ and $\angle ZYR$, it follows that the sum of the measures of $\angle XYR$ and $\angle ZYR$ is $90^{\circ }$. It's given that the measure of $\angle XYR$ is $b^{\circ }$. It follows that the measure of $\angle ZYR$ is $(90-b{)}^{\circ }$. Since $YR$ is perpendicular to $XZ$, it follows that $\angle XRY$ and $\angle ZRY$ are right angles. It follows that $∆XRY$ and $∆ZRY$ are right triangles. It's given that the measure of $\angle XZY$ is $a^{\circ }$. Because the sum of the measures of the interior angles of a triangle is $180^{\circ }$, it follows that the measures of $\angle ZRY$, $\angle XZY$, and $\angle ZYR$ have a sum of $180^{\circ }$. Since the measure of $\angle ZRY$ is $90^{\circ }$, the measure of $\angle YZR$ is $a^{\circ }$, and the measure of $\angle ZYR$ is $(90-b{)}^{\circ }$, it follows that $90+a+90-b=180$, or $a-b+180=180$. Subtracting $180$ from both sides of this equation yields $a-b=0$. Adding $b$ to both sides of this equation yields $a=b$. It's given that $\sin a^{\circ }=\frac{12}{13}$. Since $a=b$, it follows that $\sin b^{\circ }=\frac{12}{13}$. In a right triangle, the sine of an acute angle is the ratio of the length of the leg opposite the angle to the length of the hypotenuse. It follows that, for a positive constant $k$, $12k$ represents the length of the leg opposite $\angle XYR$ and $13k$ represents the length of the hypotenuse of $△XRY$. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Let $nk$ represent the length of the leg adjacent to $\angle XYR$. It follows that $(13k{)}^2=(12k{)}^2+(nk{)}^2$, or $169k^2=144k^2+n^2{k}^2$. Subtracting $144k^2$ from both sides of this equation yields $25k^2=n^2{k}^2$. Taking the positive square root of both sides of this equation yields $5k=nk$. Therefore, the length of the leg adjacent to $\angle XYR$ is $5k$. In a right triangle, the tangent of an acute angle is the ratio of the length of the leg opposite the angle to the length of the leg adjacent to the angle. It follows that $\tan b^{\circ }=\frac{12k}{5k}$, or $\tan b^{\circ }=\frac{12}{5}$. Note that 2.4 and 12/5 are examples of ways to enter a correct answer.