Right triangle RST with the right angle at T. R is at the top-left, T at the bottom-left, S at the bottom-right. Points X and Z lie on the hypotenuse RS, with X near R and Z near S. Triangle XYZ has its right angle at vertex Y, which sits outside triangle RST. Segment XY is parallel to TS and segment YZ is perpendicular to XY. In triangles RST and XYZ shown, overline{XY} is parallel to overline{TS} and tan R = dfrac{160}{231} . What is the value of sin X in triangle XYZ ? A. dfrac{160}{391} B. dfrac{160}{281} C. dfrac{231}{281} D. dfrac{231}{160}

SAT Math: Similar Right Triangles — Find sin X from tan R

A hard Digital SAT trigonometry question. Given tan R = 160/231 in right triangle RST and a parallel segment forming triangle XYZ, find sin X in triangle XYZ.

Question

Note: Figure not drawn to scale.

In triangles $RST$ and $XYZ$ shown, $\overline{XY}$ is parallel to $\overline{TS}$ and $\tan R = \dfrac{160}{231}$ . What is the value of $\sin X$ in triangle $XYZ$ ?

Step-by-Step Solution

Use the parallel sides to transfer a known angle from triangle RST into triangle XYZ.

1Read the sides of RST from tan R.

In right triangle RST the right angle is at T, so the two legs meeting at T are RT and TS, and the hypotenuse is RS. The angle at R is adjacent to RT and opposite TS.

\tan R = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{TS}{RT} = \dfrac{160}{231}

You may treat 160 and 231 as the actual side lengths — only the ratio matters for finding sin X.

2Find the hypotenuse RS with the Pythagorean theorem.

Square both legs, add, then take the square root:

RS^2 = RT^2 + TS^2 = 231^2 + 160^2 = 53{,}361 + 25{,}600 = 78{,}961

RS = \sqrt{78{,}961} = 281

So RST is a 160–231–281 right triangle. (Memorizing this triple is unnecessary — Pythagoras gets you there.)

3Use the parallel segment to match angle X to angle S.

Because X and Z both lie on segment RS, line XZ is part of line RS. That line is a transversal cutting two parallel segments: XY and TS.

The angle the transversal makes with TS at S is angle S of triangle RST. The angle it makes with XY at X is angle X of triangle XYZ. Corresponding angles formed by a transversal across parallel lines are equal, so:

\angle X \;=\; \angle S

This is the key move — it converts the unknown sin X into a sine we already know how to compute.

4Compute sin S and report sin X.

In triangle RST, the side opposite angle S is RT, and the hypotenuse is RS:

\sin S = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{RT}{RS} = \dfrac{231}{281}

Since angle X equals angle S, sin X equals sin S:

\sin X = \dfrac{231}{281}

The answer is C.