Math · Geometry and Trigonometry

SAT Right Triangles and Trigonometry: The Complete Guide

Right triangles and trigonometry questions use the Pythagorean theorem, the special right triangle ratios, and the sine, cosine, and tangent ratios to connect angles with side lengths. The same ideas extend to radians and the unit circle, where a point's coordinates are the cosine and sine of an angle. This guide gives you the complete set of facts, a method for keeping opposite and adjacent sides straight, and worked examples from easy to hard.

Section: Math, Geometry and Trigonometry
Calculator: Desmos is built in and uses radians by default; enter degrees with deg
Core facts: Pythagorean theorem, special triangles, SOHCAHTOA, radians, unit circle
Answer types: Multiple choice and student-produced response

What this skill tests

This skill covers right triangles, the sine, cosine, and tangent of acute angles, and the trigonometry the Digital SAT uses for angles in radians and points on the unit circle. Typical questions look like these:

A right triangle has two known sides. Find the third side.
In a right triangle, one acute angle has a given sine, cosine, or tangent. Find another trigonometric value.
Find a side length in a $30^\circ$ - $60^\circ$ - $90^\circ$ or $45^\circ$ - $45^\circ$ - $90^\circ$ triangle.
Convert an angle between degrees and radians, or find a coordinate on the unit circle.

Trigonometry questions often hide a complementary-angle relationship. If a sine is set equal to a cosine, check whether the two angles add to $90^\circ$ before doing anything else.

The core idea at a glance

A right triangle gives you three tools that solve most problems. The Pythagorean theorem links the three sides. The special triangles give fixed ratios. SOHCAHTOA links an acute angle to two sides. Learn the common triples so you can skip the algebra, and label each triangle relative to the angle you are working with.

Trigonometry extends the same ideas to any angle. On the unit circle, a point at angle $\theta$ , measured counterclockwise from the positive $x$ -axis, has coordinates $(\cos\theta,\sin\theta)$ . That single fact connects the ratios, the quadrant signs, and the radian measures.

Core rules, ratios, and conversions

Pythagorean theorem. For a right triangle with legs $a$ and $b$ and hypotenuse $c$ :

a^2+b^2=c^2

Common Pythagorean triples and some multiples. Each row is a complete right triangle.
Triple	Multiples
$3$ - $4$ - $5$	$6$ - $8$ - $10$ , $9$ - $12$ - $15$ , $12$ - $16$ - $20$
$5$ - $12$ - $13$	$10$ - $24$ - $26$ , $15$ - $36$ - $39$
$8$ - $15$ - $17$	$16$ - $30$ - $34$
$7$ - $24$ - $25$	$14$ - $48$ - $50$

Special right triangles. In a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle with leg $x$ , the hypotenuse is $x\sqrt2$ . In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the short leg, which faces the $30^\circ$ angle, is $x$ . The long leg, which faces the $60^\circ$ angle, is $x\sqrt3$ , and the hypotenuse is $2x$ .

x,\ x,\ x\sqrt{2}

The side lengths of a

45^\circ

45^\circ

90^\circ

triangle.

x,\ x\sqrt{3},\ 2x

The side lengths of a

30^\circ

60^\circ

90^\circ

triangle, in the order short leg, long leg, hypotenuse.

SOHCAHTOA. For an acute angle $\theta$ in a right triangle, the trigonometric ratios are defined by the sides.

\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\qquad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\qquad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}

Complementary angles. The two acute angles of a right triangle add to $90^\circ$ . So for any acute angle $x$ , the sine of $x$ equals the cosine of its complement, and the reverse holds too.

\sin x=\cos(90^\circ-x),\qquad \cos x=\sin(90^\circ-x)

Special values you should know without a calculator are in the table below. The reference sheet includes the special triangle ratios, but these values appear often enough that memorizing them saves time.

Angle	sin	cos	tan
$30^\circ$	$\frac12$	$\frac{\sqrt3}{2}$	$\frac{\sqrt3}{3}$
$45^\circ$	$\frac{\sqrt2}{2}$	$\frac{\sqrt2}{2}$	$1$
$60^\circ$	$\frac{\sqrt3}{2}$	$\frac12$	$\sqrt3$

Radians. A full turn is $360^\circ$ , or $2\pi$ radians, so $180^\circ=\pi$ radians. To convert, multiply by $\frac{\pi}{180}$ to go from degrees to radians, or by $\frac{180}{\pi}$ to go back.

\text{radians}=\text{degrees}\cdot\frac{\pi}{180}\qquad\text{degrees}=\text{radians}\cdot\frac{180}{\pi}

Degrees	Radians
$30^\circ$	$\frac{\pi}{6}$
$45^\circ$	$\frac{\pi}{4}$
$60^\circ$	$\frac{\pi}{3}$
$90^\circ$	$\frac{\pi}{2}$
$120^\circ$	$\frac{2\pi}{3}$
$150^\circ$	$\frac{5\pi}{6}$
$180^\circ$	$\pi$
$270^\circ$	$\frac{3\pi}{2}$
$360^\circ$	$2\pi$

The unit circle. A unit circle has radius $1$ and is centered at the origin, so every point $(x,y)$ on it satisfies $x^2+y^2=1$ . For an angle $\theta$ , the point where its terminal side meets the circle is $(\cos\theta,\sin\theta)$ . The signs of the coordinates depend on the quadrant.

Quadrant signs. A common memory cue is All Students Take Calculus, starting from Quadrant I.
Quadrant	Angle range	Sine	Cosine	Tangent
I	$0$ to $\frac{\pi}{2}$	+	+	+
II	$\frac{\pi}{2}$ to $\pi$	+	−	−
III	$\pi$ to $\frac{3\pi}{2}$	−	−	+
IV	$\frac{3\pi}{2}$ to $2\pi$	−	+	−

Step-by-step method

Locate the right angle and the hypotenuse, which is the side across from the right angle.
Choose the acute angle that the question mentions, and label the other two sides as opposite and adjacent to that angle.
Pick the ratio that links the known values to the unknown: SOH for sine, CAH for cosine, and TOA for tangent.
If a side is missing, use the Pythagorean theorem or a special-triangle ratio. If the sides form a known triple, the answer may be immediate.
For angles, convert radians to degrees, or the reverse, before you compare. Then check the quadrant and the sign.
Sanity check: the hypotenuse must be the longest side, and sine and cosine must lie between $-1$ and $1$ .

Worked examples

The first example uses only the Pythagorean theorem, the second uses a complementary angle, and the third asks for a trigonometric value in a quadrant where the sign is the main difficulty.

Worked example

Example 1: A missing hypotenuse (easy)

Note: Figure not drawn to scale.

The right triangle shown has legs of $6$ units and $8$ units. What is the length of its hypotenuse, in units?

A $10$
B $14$
C $48$
D $100$

Show the solution

The hypotenuse is across from the right angle, so it is the side $c$ in $a^2+b^2=c^2$ .
Substitute: $6^2+8^2=c^2$ , so $36+64=c^2$ .
Add to get $100=c^2$ . Take the positive square root to get $c=10$ . The answer is A.
Why not B ( $14$ )? That is $6+8$ . Adding the legs does not give the hypotenuse.
Why not C ( $48$ )? That is $6\cdot8$ , the product of the legs. It is twice the area of the triangle, which is $24$ .
Why not D ( $100$ )? That is $c^2$ , not $c$ . The question asks for the length, so the square root is required.

Answer: A

Worked example

Example 2: A complementary cosine (medium)

In right triangle $ABC$ , the right angle is at $C$ . If $\sin A=\frac35$ , what is $\cos B$ ?

A $\frac45$
B $\frac35$
C $\frac53$
D $\frac34$

Show the solution

In a right triangle, $\angle A$ and $\angle B$ are complementary, so $\angle B=90^\circ-\angle A$ .
Apply the complement identity: $\cos B=\cos(90^\circ-A)=\sin A=\frac35$ .
Check with the triangle. $\sin A=\frac{BC}{AB}=\frac35$ , so we can take $BC=3$ and $AB=5$ . Then $AC=4$ . For angle $B$ , the adjacent leg is $BC=3$ and the hypotenuse is $5$ , so $\cos B=\frac35$ . The answer is B.
Why not A ( $\frac45$ )? That is $\cos A$ . It uses the leg adjacent to $A$ , which is the wrong angle.
Why not C ( $\frac53$ )? That is $\frac{1}{\sin A}$ . Cosine can never be larger than $1$ , so a value above one is impossible.
Why not D ( $\frac34$ )? That is $\tan A$ , the opposite leg over the adjacent leg for angle $A$ .

Answer: B

Worked example

Example 3: A coordinate on the unit circle (hard)

A point $P$ lies on the unit circle in Quadrant III. The angle $\theta$ from the positive $x$ -axis to the ray $\overrightarrow{OP}$ satisfies $\sin\theta=-\frac35$ . What is $\tan\theta$ ?

A $-\frac34$
B $\frac34$
C $-\frac43$
D $\frac43$

Show the solution

Use the Pythagorean identity: $\cos^2\theta=1-\sin^2\theta=1-\frac{9}{25}=\frac{16}{25}$ , so $\cos\theta=\pm\frac45$ .
Quadrant III is where both sine and cosine are negative, so $\cos\theta=-\frac45$ .
Then $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-3/5}{-4/5}=\frac34$ . The answer is B.
Why not A ( $-\frac34$ )? The sign is wrong. Tangent is positive in Quadrant III because sine and cosine are both negative, and a negative divided by a negative is positive.
Why not C ( $-\frac43$ )? This inverts the ratio, and it also carries the wrong sign. The question asks for sine over cosine.
Why not D ( $\frac43$ )? This is $\cos\theta$ divided by $\sin\theta$ , the reciprocal of the tangent. Check which ratio the question defines.

Answer: B

Common traps

The hardest questions

The harder questions combine a right triangle with an identity, a radian angle, or a real-world setting. These four patterns cover most of them.

Pattern 1: complementary equations. If $\sin(3x+10^\circ)=\cos(x+20^\circ)$ , the two angles are complementary in a right-triangle setting. Set $(3x+10)+(x+20)=90$ , which gives $4x=60$ and $x=15$ .

Pattern 2: a ladder or a height. A $10$ -foot ladder leans against a wall and makes a $60^\circ$ angle with the ground. The height it reaches on the wall is $10\sin60^\circ=10\cdot\frac{\sqrt3}{2}=5\sqrt3\approx8.66$ feet.

Pattern 3: coordinates and distance. The point $(-3,4)$ is $\sqrt{(-3)^2+4^2}=5$ units from the origin. Scaling by $5$ gives a point on the unit circle at the same angle, with coordinates $\left(-\frac35,\frac45\right)$ . Dividing each coordinate by the distance is the key step.

Pattern 4: special angles beyond Quadrant I. The point at $\frac{2\pi}{3}$ has reference angle $\pi-\frac{2\pi}{3}=\frac{\pi}{3}$ . Quadrant II has a negative cosine, so the coordinates are $\left(-\frac12,\frac{\sqrt3}{2}\right)$ . Find the reference angle first, then apply the signs.

Timing and practice plan

A labeled right triangle takes under a minute, and a multi-step identity or unit-circle question can take two or three minutes. The fastest students recognize the common triples, so drill those until you can spot them in a stem.

Write the four triples and their multiples from memory, then check yourself against the table in this guide.
Practice labeling: take ten right triangles and write the opposite, adjacent, and hypotenuse for each acute angle before calculating.
Memorize the special values table, and time yourself until you can fill it in under a minute.
Solve five complementary-angle questions and five unit-circle questions that require a quadrant sign.
Convert ten angles between degrees and radians in both directions.
Review each error and label it as a labeling error, a sign error, a mode error, or a memory gap.

In Solar, use this skill's practice questions for the drills above. Move on once you can solve a mixed set without a sign or labeling error.

Keep reading — it’s free

Create a free Solar account to unlock the rest of this right triangles and trigonometry guide, plus adaptive practice on exactly this skill.

Core rules, ratios, and conversions
Step-by-step method
Worked examples
Common traps
The hardest questions

Unlock the full guide Already have an account? Log in

Right triangles and trigonometry: FAQ

Which right triangle ratios should I memorize?

Memorize the 3-4-5, 5-12-13, 8-15-17, and 7-24-25 triples and their multiples, along with the 45-45-90 and 30-60-90 side ratios. The reference sheet includes the special triangle ratios, but knowing the triples saves time.

Do I need to know radians for the SAT?

Yes. The Digital SAT uses radians in some trigonometry and circle questions, and converting between degrees and radians is a common step. Memorize that 180 degrees equals pi radians.

Is sin x equal to cos(90° − x)?

Yes. In a right triangle, the two acute angles are complementary, so the sine of one equals the cosine of the other. The identity holds for any acute angle x.

Can Desmos handle trigonometry on the SAT?

Yes. Desmos is built into the Math section. It works in radians by default, so enter degrees with the degree symbol or the word deg, and check the mode before you trust a value.

What is the unit circle used for on the SAT?

Mostly to link an angle to coordinates: a point at angle θ on the unit circle is (cos θ, sin θ). Knowing the quadrant signs and the standard angles covers most questions.