A right-triangular sail is shaped like triangle , with a right angle at . Point lies on such that feet and feet. A seam is stitched from to a point on so that is parallel to . If feet, which choice is the value of ?

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SAT Right Triangles & Trigonometry Question 30 | Free SAT Prep | Aniko

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Question 30·Hard·Right Triangles and Trigonometry

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A right-triangular sail is shaped like triangle $ABC$ , with a right angle at $C$ . Point $E$ lies on $\overline{BC}$ such that $CE=16$ feet and $EB=24$ feet.

A seam is stitched from $E$ to a point $D$ on $\overline{AC}$ so that $\overline{DE}$ is parallel to $\overline{AB}$ . If $AD=15$ feet, which choice is the value of $\tan(\angle A)$ ?

When a segment inside a triangle is parallel to one side, expect similar triangles. Write a proportion using corresponding sides to connect a smaller segment length to a full side length, solve for the needed leg, and then apply $\tan(\text{angle})=\frac{\text{opposite}}{\text{adjacent}}$ using the legs of the right triangle.

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Use the segment lengths on $BC$

First find $BC$ by adding the two given parts: $CE$ and $EB$ .

Look for similar triangles

Since $DE$ is parallel to $AB$ , the small triangle with vertex at $C$ is similar to the whole triangle.

Relate a part of $AC$ to the whole $AC$

Let $CD=x$ , so $CA=x+15$ . Use a proportion like $\frac{CE}{CB}=\frac{CD}{CA}$ to solve for $x$ .

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Compute $BC$

Compute $BC=16+24$ .

Solve for $CD$ with a proportion

Let $x$ represent $CD$ . Enter the equation 16/40 = x/(x+15) and solve for $x$ .

Compute the tangent

Compute $\tan(\angle A)=BC/(x+15)$ using $BC=40$ and your solved value of $x$ .

Step-by-step Explanation

Find the full length of $BC$

Since $E$ lies on $\overline{BC}$ and $CE=16$ and $EB=24$ ,

BC = CE+EB = 16+24 = 40.

Use similarity from the parallel seam

Because $DE\parallel AB$ , triangles $\triangle CDE$ and $\triangle CBA$ are similar (they share $\angle C$ , and corresponding angles formed by a parallel line are equal).

So corresponding side ratios satisfy

\frac{CE}{CB}=\frac{CD}{CA}.

Solve for $CA$ using a variable

Let $CD=x$ . Then $CA = AD+CD = 15+x$ .

Using the similarity ratio:

\frac{16}{40}=\frac{x}{15+x} \Rightarrow \frac{2}{5}=\frac{x}{15+x}.

Solve:

5x = 2(15+x) \Rightarrow 5x=30+2x \Rightarrow 3x=30 \Rightarrow x=10.

Thus $CA=15+10=25$ .

Compute $\tan(\angle A)$

In right triangle $ABC$ (right angle at $C$ ),

\tan(\angle A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AC}=\frac{40}{25} = \frac{8}{5}.

So the correct choice is $\frac{8}{5}$ .

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