SAT Question 2757549e: Right Triangles And Trigonometry answer and explanation with Desmos solution

Choice C is correct. It’s given that the length of $\overline{DE}$ is $147$, and the length of $\overline{AE}$ is $245$. Since angle $D$ is a right angle, triangle $ADE$ is a right triangle with hypotenuse $\overline{AE}$ and legs $\overline{AD}$ and $\overline{DE}$. By the Pythagorean theorem, if a right triangle has a hypotenuse with length $c$ and legs with lengths $a$ and $b$, then $a^2+b^2=c^2$. Substituting $AD$ for $a$, $147$ for $b$, and $245$ for $c$ in the equation $a^2+b^2=c^2$ yields $A{D}^2+147^2=245^2$. Subtracting $147^2$ from both sides of this equation yields $A{D}^2=245^2-147^2$. Evaluating this difference yields $A{D}^2=60,025-21,609$, or $A{D}^2=38,416$. Taking the positive square root of both sides of this equation yields $AD=196$. It's given that $\overline{BC}$ is parallel to $\overline{DE}$, and $\overline{AE}$ intersects $\overline{BC}$ and $\overline{DE}$. It follows that angle $ACB$ is congruent to angle $AED$. Since angle $A$ is common to triangles $ABC$ and $ADE$, it follows that triangle $ABC$ is similar to triangle $ADE$. In triangle $ABC$, $\tan x^{\circ }$ is the ratio of the leg opposite angle $ACB$ to the leg adjacent to angle $ACB$, so $\tan x^{\circ }=\frac{AB}{BC}$. Since triangle $ABC$ is similar to triangle $ADE$ and similar triangles have proportional side lengths, it follows that $\frac{AB}{BC}=\frac{AD}{DE}$. Therefore, $\tan x^{\circ }=\frac{AD}{DE}$. Substituting $196$ for $AD$ and $147$ for $DE$ in this equation yields $\tan x^{\circ }=\frac{196}{147}$. Therefore, the value of $\tan x^{\circ }$ is $\frac{196}{147}$.