# Right triangles and trigonometry on the SAT Canonical: https://duckyhelper.com/learn/sat-math/right-triangles-and-trigonometry/ Updated: 2026-10-01 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 \(\sin x^\circ = \cos(90 - x)^\circ\). Radians also appear: \(180^\circ = \pi\) radians. ## The key ideas $$ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan A = \frac{\text{opposite}}{\text{adjacent}} $$ "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)** | Triangle | Side lengths | Example | | --- | --- | --- | | 45-45-90 | \(s, s, s\sqrt{2}\) | Legs 5 and 5, hypotenuse \(5\sqrt{2}\) | | 30-60-90 | \(x, x\sqrt{3}, 2x\) | Short leg 4, long leg \(4\sqrt{3}\), hypotenuse 8 | The two acute angles of a right triangle add to \(90^\circ\), and the side opposite one is adjacent to the other. That is why \(\sin x^\circ = \cos(90 - x)^\circ\). $$ \text{radians} = \text{degrees} \times \frac{\pi}{180} $$ ## 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. $$ \sqrt{26^2 - 10^2} = \sqrt{576} = 24 $$ 2. Tangent is opposite over adjacent. The opposite side is 10 and the adjacent leg is 24. $$ \frac{10}{24} = \frac{5}{12} $$ Answer: \(\frac{5}{12}\) **Example 2: from one ratio to the others** Problem: In right triangle ABC with the right angle at C, \(\sin A = \frac{8}{17}\). Find \(\cos A\) and \(\tan A\). 1. Think of the opposite side as 8 and the hypotenuse as 17. Find the adjacent side. $$ \sqrt{17^2 - 8^2} = \sqrt{225} = 15 $$ 2. Cosine is adjacent over hypotenuse. $$ \cos A = \frac{15}{17} $$ 3. Tangent is opposite over adjacent. $$ \tan A = \frac{8}{15} $$ Answer: \(\cos A = \frac{15}{17}\) and \(\tan A = \frac{8}{15}\) **Example 3: complementary angles** Problem: If \(\sin(2x + 10)^\circ = \cos(3x - 5)^\circ\), 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. $$ (2x + 10) + (3x - 5) = 90 $$ 2. Combine like terms. $$ 5x + 5 = 90 $$ 3. Solve. $$ x = 17 $$ 4. Check: the angles are \(44^\circ\) and \(46^\circ\), which add to \(90^\circ\). Answer: \(x = 17\) **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 \(\sqrt{3}\). $$ h = 5\sqrt{3} $$ 3. Area is one half base times height. $$ \frac{1}{2}(10)(5\sqrt{3}) = 25\sqrt{3} $$ Answer: \(25\sqrt{3}\), 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: \(26^2 - 10^2\), not \(26^2 + 10^2\). Fix: the hypotenuse squared is the biggest number. - **Mixing up the 30-60-90 sides.** The side \(x\sqrt{3}\) is across from \(60^\circ\), and \(2x\) is the hypotenuse. Fix: the shortest side is across from the smallest angle. - **Setting complementary angles equal.** \(\sin a = \cos b\) means \(a + b = 90\), not \(a = b\). - **Calculator in the wrong mode.** \(\sin(30)\) in radian mode is about \(-0.99\). Fix: check the degree or radian setting in Desmos before using trig keys. ## Quick method > **Tip: Know the common triples** > > 3-4-5, 5-12-13, 8-15-17 and 7-24-25 (and their multiples, like 10-24-26) show up again and again. Spotting one saves the square root step. This is a real shortcut, but check that the numbers really match a triple before you use it. ## 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? A. \(\frac{7}{24}\) B. \(\frac{7}{25}\) C. \(\frac{24}{25}\) D. \(\frac{25}{7}\) Answer: \(\frac{7}{25}\). The hypotenuse is \(\sqrt{49 + 576} = 25\). Sine is opposite over hypotenuse: \(\frac{7}{25}\). \(\frac{7}{24}\) is the tangent, and \(\frac{24}{25}\) is the cosine. 2. If \(\sin x^\circ = \cos 32^\circ\) and x is between 0 and 90, what is x? A. 32 B. 58 C. 68 D. 148 Answer: 58. Complementary angles: \(x = 90 - 32 = 58\). 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? A. \(5\) B. \(5\sqrt{2}\) C. \(10\sqrt{2}\) D. \(5\sqrt{3}\) Answer: \(5\sqrt{2}\). Leg times \(\sqrt{2}\) is the hypotenuse, so leg is \(\frac{10}{\sqrt{2}} = 5\sqrt{2}\). \(10\sqrt{2}\) multiplies instead of dividing. \(5\sqrt{3}\) is from a 30-60-90 triangle. 4. An angle measures \(\frac{3\pi}{4}\) radians. What is its measure in degrees? A. 45 B. 135 C. 225 D. 270 Answer: 135. Multiply by \(\frac{180}{\pi}\): \(\frac{3\pi}{4} \cdot \frac{180}{\pi} = 135\). 45 is \(\frac{\pi}{4}\), and 225 is \(\frac{5\pi}{4}\). 5. Student-produced response: in right triangle PQR, the right angle is at Q, \(\cos P = 0.6\) and \(PR = 15\). What is the length of QR? Answer: 12. Cosine is adjacent over hypotenuse, so \(PQ = 0.6 \times 15 = 9\). Then \(QR = \sqrt{15^2 - 9^2} = \sqrt{144} = 12\). 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 \(90^\circ\). 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 \(2\pi\) radians, of arc. You should convert between degrees and radians and know common values like \(\frac{\pi}{6} = 30^\circ\) and \(\frac{\pi}{2} = 90^\circ\). ## Sources - [College Board: SAT Math, Geometry and Trigonometry skills](https://satsuite.collegeboard.org/sat/whats-on-the-test/math/types/geometry-trigonometry), accessed 2026-10-01 - [College Board: SAT Bluebook test directions and reference sheet](https://satsuite.collegeboard.org/media/pdf/english-sat-test-directions-bb.pdf), accessed 2026-10-01 ## Related - [Lines, angles and triangles on the SAT](https://duckyhelper.com/learn/sat-math/lines-angles-and-triangles/) - [Circles on the SAT](https://duckyhelper.com/learn/sat-math/circles/) - [Special right triangles: 45-45-90 and 30-60-90](https://duckyhelper.com/learn/geometry/special-right-triangles/) - [Right triangle trigonometry and SOHCAHTOA](https://duckyhelper.com/learn/geometry/right-triangle-trigonometry/) - [SAT Math study guides](https://duckyhelper.com/learn/sat-math/) ## 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." ## Get DuckyHelper Free to start. 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