SAT Question 8027db3f: Right Triangles And Trigonometry answer and explanation with Desmos solution

It's given that angle $J$ is the right angle in triangle $JKL$. Therefore, the acute angles of triangle $JKL$ are angle $K$ and angle $L$. The hypotenuse of a right triangle is the side opposite its right angle. Therefore, the hypotenuse of triangle $JKL$ is side $KL$. The cosine of an acute angle in a right triangle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. It's given that $\cos \left(K\right)=\frac{24}{51}$. This can be written as $\cos \left(K\right)=\frac{8}{17}$. Since the cosine of angle $K$ is a ratio, it follows that the length of the side adjacent to angle $K$ is $8n$ and the length of the hypotenuse is $17n$, where $n$ is a constant. Therefore, $JK=8n$ and $KL=17n$. 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. For triangle $JKL$, it follows that ${\left(JK\right)}^2+{\left(JL\right)}^2={\left(KL\right)}^2$. Substituting $8n$ for $JK$ and $17n$ for $KL$ yields ${\left(8n\right)}^2+{\left(JL\right)}^2={\left(17n\right)}^2$. This is equivalent to $64n^2+{\left(JL\right)}^2=289n^2$. Subtracting $64n^2$ from each side of this equation yields ${\left(JL\right)}^2=225n^2$. Taking the square root of each side of this equation yields $JL=15n$. Since $\cos \left(L\right)=\frac{JL}{KL}$, it follows that $\cos \left(L\right)=\frac{15n}{17n}$, which can be rewritten as $\cos \left(L\right)=\frac{15}{17}$. Note that 15/17, .8823, .8824, and 0.882 are examples of ways to enter a correct answer.