SAT Math · Geometry and Trigonometry

Right triangles and trigonometry on the SAT

SAT right triangle questions use the Pythagorean theorem, the special 45-45-90 and 30-60-90 triangles, and the trig ratios: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, tangent is opposite over adjacent. A favorite fact: the sine of one acute angle equals the cosine of the other, so sin⁡x∘=cos⁡(90−x)∘\sin x^\circ = \cos(90 - x)^\circ. Radians also appear: 180∘=π180^\circ = \pi radians.

Updated

On this page

The key ideas

sin⁡A=oppositehypotenusecos⁡A=adjacenthypotenusetan⁡A=oppositeadjacent\sin A = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan A = \frac{\text{opposite}}{\text{adjacent}}

"Opposite" and "adjacent" depend on which angle you are standing at. The hypotenuse is always the side across from the right angle, and it is the longest side.

Special right triangles (also on the SAT reference sheet)
TriangleSide lengthsExample
45-45-90s,s,s2s, s, s\sqrt{2}Legs 5 and 5, hypotenuse 525\sqrt{2}
30-60-90x,x3,2xx, x\sqrt{3}, 2xShort leg 4, long leg 434\sqrt{3}, hypotenuse 8

The two acute angles of a right triangle add to 90∘90^\circ, and the side opposite one is adjacent to the other. That is why sin⁡x∘=cos⁡(90−x)∘\sin x^\circ = \cos(90 - x)^\circ.

radians=degrees×π180\text{radians} = \text{degrees} \times \frac{\pi}{180}

Worked examples

Example 1: find a side, then a ratio

Problem A right triangle has a hypotenuse of 26 and one leg of 10. What is the tangent of the angle opposite the leg of length 10?

  1. Find the other leg with the Pythagorean theorem.
    262−102=576=24\sqrt{26^2 - 10^2} = \sqrt{576} = 24
  2. Tangent is opposite over adjacent. The opposite side is 10 and the adjacent leg is 24.
    1024=512\frac{10}{24} = \frac{5}{12}

Answer 512\frac{5}{12}

Example 2: from one ratio to the others

Problem In right triangle ABC with the right angle at C, sin⁡A=817\sin A = \frac{8}{17}. Find cos⁡A\cos A and tan⁡A\tan A.

  1. Think of the opposite side as 8 and the hypotenuse as 17. Find the adjacent side.
    172−82=225=15\sqrt{17^2 - 8^2} = \sqrt{225} = 15
  2. Cosine is adjacent over hypotenuse.
    cos⁡A=1517\cos A = \frac{15}{17}
  3. Tangent is opposite over adjacent.
    tan⁡A=815\tan A = \frac{8}{15}

Answer cos⁡A=1517\cos A = \frac{15}{17} and tan⁡A=815\tan A = \frac{8}{15}

Example 3: complementary angles

Problem If sin⁡(2x+10)∘=cos⁡(3x−5)∘\sin(2x + 10)^\circ = \cos(3x - 5)^\circ, and both angles are acute, what is x?

  1. Sine of an angle equals cosine of its complement, so the two angles add to 90.
    (2x+10)+(3x−5)=90(2x + 10) + (3x - 5) = 90
  2. Combine like terms.
    5x+5=905x + 5 = 90
  3. Solve.
    x=17x = 17
  4. Check: the angles are 44∘44^\circ and 46∘46^\circ, which add to 90∘90^\circ.

Answer x=17x = 17

Example 4 (SAT-hard): a 30-60-90 inside an equilateral triangle

Problem An equilateral triangle has sides of length 10. What is its area?

  1. Draw the height from one vertex. It splits the triangle into two 30-60-90 triangles with hypotenuse 10 and short leg 5.
  2. The long leg (the height) is the short leg times 3\sqrt{3}.
    h=53h = 5\sqrt{3}
  3. Area is one half base times height.
    12(10)(53)=253\frac{1}{2}(10)(5\sqrt{3}) = 25\sqrt{3}

Answer 25325\sqrt{3}, about 43.3

Common mistakes

  • Using the wrong angle's sides. Opposite and adjacent swap when you move to the other acute angle. Fix: put your finger on the angle first, then name the sides.
  • Adding squares for a leg. To find a leg, subtract: 262−10226^2 - 10^2, not 262+10226^2 + 10^2. Fix: the hypotenuse squared is the biggest number.
  • Mixing up the 30-60-90 sides. The side x3x\sqrt{3} is across from 60∘60^\circ, and 2x2x is the hypotenuse. Fix: the shortest side is across from the smallest angle.
  • Setting complementary angles equal. sin⁡a=cos⁡b\sin a = \cos b means a+b=90a + b = 90, not a=ba = b.
  • Calculator in the wrong mode. sin⁡(30)\sin(30) in radian mode is about −0.99-0.99. Fix: check the degree or radian setting in Desmos before using trig keys.

Quick method

Practice

5 SAT-style questions

  1. A right triangle has legs of 7 and 24. What is the sine of the angle opposite the side of length 7?

    1. 724\frac{7}{24}
    2. 725\frac{7}{25}
    3. 2425\frac{24}{25}
    4. 257\frac{25}{7}
    Show answer

    Answer: 725\frac{7}{25}

    The hypotenuse is 49+576=25\sqrt{49 + 576} = 25. Sine is opposite over hypotenuse: 725\frac{7}{25}. 724\frac{7}{24} is the tangent, and 2425\frac{24}{25} is the cosine.

  2. If sin⁡x∘=cos⁡32∘\sin x^\circ = \cos 32^\circ and x is between 0 and 90, what is x?

    1. 32
    2. 58
    3. 68
    4. 148
    Show answer

    Answer: 58

    Complementary angles: x=90−32=58x = 90 - 32 = 58. 32 sets the angles equal, and 148 uses 180 instead of 90.

  3. A 45-45-90 triangle has a hypotenuse of 10. How long is each leg?

    1. 55
    2. 525\sqrt{2}
    3. 10210\sqrt{2}
    4. 535\sqrt{3}
    Show answer

    Answer: 525\sqrt{2}

    Leg times 2\sqrt{2} is the hypotenuse, so leg is 102=52\frac{10}{\sqrt{2}} = 5\sqrt{2}. 10210\sqrt{2} multiplies instead of dividing. 535\sqrt{3} is from a 30-60-90 triangle.

  4. An angle measures 3π4\frac{3\pi}{4} radians. What is its measure in degrees?

    1. 45
    2. 135
    3. 225
    4. 270
    Show answer

    Answer: 135

    Multiply by 180π\frac{180}{\pi}: 3π4⋅180π=135\frac{3\pi}{4} \cdot \frac{180}{\pi} = 135. 45 is π4\frac{\pi}{4}, and 225 is 5π4\frac{5\pi}{4}.

  5. Student-produced response: in right triangle PQR, the right angle is at Q, cos⁡P=0.6\cos P = 0.6 and PR=15PR = 15. What is the length of QR?

    Show answer

    Answer: 12

    Cosine is adjacent over hypotenuse, so PQ=0.6×15=9PQ = 0.6 \times 15 = 9. Then QR=152−92=144=12QR = \sqrt{15^2 - 9^2} = \sqrt{144} = 12. It is a 3-4-5 triangle scaled by 3.

Frequently asked questions

What does SOHCAHTOA mean?

It is a memory trick for the three ratios: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. Always name the sides from the angle you are working with.

Why is sin x equal to cos(90 - x)?

In a right triangle the two acute angles add to 90∘90^\circ. The side opposite one angle is the side adjacent to the other, so the sine of one equals the cosine of the other. SAT questions use this to set up an equation.

Are radians on the SAT?

Yes. College Board's reference sheet notes that a circle has 360 degrees, or 2π2\pi radians, of arc. You should convert between degrees and radians and know common values like π6=30∘\frac{\pi}{6} = 30^\circ and π2=90∘\frac{\pi}{2} = 90^\circ.

Sources

  1. College Board: SAT Math, Geometry and Trigonometry skills, accessed October 1, 2026
  2. College Board: SAT Bluebook test directions and reference sheet, accessed October 1, 2026

Try asking Ducky

  • “Which side is opposite and which is adjacent here?”
  • “Why does sin 32 equal cos 58?”
  • “Quiz me on the special right triangles.”

Free to start. The web app works in any browser, Chromebooks included; the Mac app can also draw on your real screen.

Try it free in your browserGet the Mac app
DuckyHelper
The AI tutor that shows you
FREE TO TRY
Try it free