Question 520·Hard·Math

Because the parallel seam $DE$ runs alongside $AB$, the angle at $D$ equals the angle at $A$. Both triangles also have the right angle at $C$. Two equal pairs of angles make triangles similar, meaning the same shape: $\triangle CDE\sim\triangle CAB$. So $C$ matches $C$, $D$ matches $A$, and $E$ matches $B$.

Point $E$ lies between $C$ and $B$, so the whole side $CB$ contains both labeled pieces. Add them:

Let $x=CD$. Point $D$ lies between $A$ and $C$, so $AC=x+15$. The similar triangles pair $CD$ with the whole side $CA$, and $CE$ with the whole side $CB$. Keep small over whole in both ratios: $\frac{x}{x+15}=\frac{16}{40}$. Pairing $CD$ with $AD=15$ would use a leftover piece instead of its matching side.

Type $\frac{x_1}{x_1+15}\sim\frac{16}{40}$. The subscript makes $x_1$ a value Desmos can find, and $\sim$ tells Desmos to make the two sides equal. Under PARAMETERS, it reports $x_1=10$, so the missing leg $CD$ is $10$ feet.

The angle at $D$ matches $\angle A$. In the smaller right triangle, $CE$ is opposite angle $D$ and $CD$ is the adjacent leg beside it. Tangent is opposite over adjacent, so: $\tan(\angle A)=\tan(\angle D)=\frac{CE}{CD}=\frac{16}{10}$. Reduce the fraction: $\tan(\angle A)=\frac{8}{5}$. For the sail’s angle at $A$, tangent is $\frac{8}{5}$. Choice A.