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Original practice · Math

SAT Trigonometry: Read Sine from a Right Triangle

Identify opposite, adjacent and hypotenuse in an original labeled right triangle, then calculate sine.

Reviewed

Before you start

Practice right triangles and trigonometry with an original problem, a complete solution and a transfer check. Try the problem before reading the worked answer.

The original problem

Right triangle ABC has a right angle at B. AB = 8 and BC = 15. What is sin(A)?

Right triangle ABC has A at lower left, B at lower right with a right-angle marker, and C above B. AB is 8, BC is 15 and AC is 17.
Try the question before reading the solution

Choose one answer, then check it.

Answer: 15/17

Find the hypotenuse

AC is opposite the right angle. AC² = 8² + 15² = 64 + 225 = 289, so AC = 17.

Identify the side opposite angle A

BC, of length 15, is opposite angle A. AB, of length 8, is adjacent to A.

Apply the sine ratio

sin(A) = opposite/hypotenuse = BC/AC = 15/17.

A second method or independent check

Angle C is complementary to angle A. Its cosine uses its adjacent leg BC and the same hypotenuse AC: cos(C) = 15/17. The identity sin(A) = cos(C) provides a second check.

Why tempting answers fail

  • 8/17 is cos(A), not sin(A).
  • 8/15 is the reciprocal of tan(A).
  • 15/8 is tan(A). Sine of this acute angle must be between 0 and 1.

Try a transfer problem

What is cos(A) for the same triangle?

Answer: 8/17

Cosine is adjacent/hypotenuse. Relative to A, AB is the adjacent leg, so cos(A) = 8/17.

What to practice next

If this was difficult, return to right triangles and trigonometry practice and identify the precise step to improve. These are original PeakSAT teaching examples, not official College Board questions or calibrated score predictions. For official adaptive test practice, use Bluebook (opens in a new tab).

Sources

  1. College Board: SAT Math contentsatsuite.collegeboard.org

Sources are linked for reference; they do not endorse PeakSAT. How we prepare our guides