SAT Question c6dff223: Right Triangles And Trigonometry answer and explanation with Desmos solution

It's given that triangle $ABC$ is similar to triangle $DEF$, where angle $A$ corresponds to angle $D$ and angles $C$ and $F$ are right angles. In similar triangles, the tangents of corresponding angles are equal. Therefore, if $\tan A=\frac{21}{20}$, then $\tan D=\frac{21}{20}$. 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. Therefore, in triangle $DEF$, if $\tan D=\frac{21}{20}$, the ratio of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ to the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $\frac{21}{20}$. If the lengths of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ and $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ are $21$ and $20$, respectively, then the ratio of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ to the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $\frac{21}{20}$. 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. Therefore, the value of $\sin D$ is the ratio of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ to the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$. The length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ can be calculated using the Pythagorean theorem, which states that if the lengths of the legs of a right triangle are $a$ and $b$ and the length of the hypotenuse is $c$, then $a^2+b^2=c^2$. Therefore, if the lengths of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ and $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ are $21$ and $20$, respectively, then ${\left(21\right)}^2+{\left(20\right)}^2={\left(DE\right)}^2$, or $841={\left(DE\right)}^2$. Taking the positive square root of both sides of this equation yields $29=DE$. Therefore, if the lengths of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ and $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ are $21$ and $20$, respectively, then the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $29$ and the ratio of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ to the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $\frac{21}{29}$. Thus, if $\tan A=\frac{21}{20}$, the value of $\sin D$ is $\frac{21}{29}$. Note that 21/29, .7241, and 0.724 are examples of ways to enter a correct answer.