SAT Question c4a9c94d: Right Triangles And Trigonometry answer and explanation with Desmos solution

5$.Thefigureshowsthat$\triangle B A C$isarighttriangle.Inthefigure,$\overline{A D}$isperpendicularto$\overline{B C}$,whichformstwoadditionalrighttriangles:$\triangle B D A$and$\triangle A D C$.Iftwopairsofcorrespondinganglesintwotrianglesarecongruent,thenthetrianglesaresimilar.$\triangle B D A$sharesangle$B$with$\triangle B A C$and$\triangle A D C$sharesangle$C$with$\triangle B A C$.Therefore,thefigurecontainsthreesimilarrighttriangles:$\triangle B A C$,$\triangle B D A$,and$\triangle A D C$.Insimilartriangles,thetangentsofcorrespondinganglesareequivalent.I{t}^{\prime }sgiventhatthetangentofangle$C A D$is$1.3$.Sinceangle$C A D$in$\triangle A D C$correspondstoangle$A B D$in$\triangle B D A$,itfollowsthatthetangentofangle$A B D$isalso＄1.3$. The tangent of an acute angle in a right triangle is equal to the length of the leg opposite the angle divided by the length of the leg adjacent to the angle. Therefore, $\tan ABD=\frac{AD}{BD}$. It's given that the length of $\overline{BD}$ is $5$. Substituting ＄5$for$B D$and$1.3$for$\tan A B D$intheequation$\tan A B D = \frac{A D}{B D}$yields$1.3 = \frac{A D}{5}$.Multiplyingbothsidesofthisequationby＄5$ yields $6.5=AD$. Therefore, the length of $\overline{AD}$ is $6.5$. Note that 6.5 and 13/2 are examples of ways to enter a correct answer.