Right Triangle Trigonometry

Angles In, Side Ratios Out

Right-triangle trigonometry is a machine that turns an angle into a ratio of sides and back again. On the SAT you use it in three ways: find a missing side when you know an angle, find a missing angle when you know two sides, and exploit a slick relationship between sine and cosine. Everything on the SAT stays inside a single right triangle — there is no law of sines or law of cosines to learn here.

Naming the Sides From the Angle's Point of View

Pick one acute angle and call it $\theta$. Relative to $\theta$, the three sides get names:

• the opposite side is directly across the triangle from $\theta$;
• the adjacent side is next to $\theta$ but is not the hypotenuse;
• the hypotenuse is across from the right angle and is always the longest side.

The word "adjacent" causes most of the trouble, because the hypotenuse also touches $\theta$. The rule is fixed: the hypotenuse is never called the adjacent side. And crucially, "opposite" and "adjacent" swap when you move to the other acute angle — the side opposite one acute angle is adjacent to the other.

SOH-CAH-TOA

The three primary ratios are captured by the mnemonic SOH-CAH-TOA:

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

To choose the right ratio, list what you have and what you want. If you have the hypotenuse and want the opposite side, the pair "opposite + hypotenuse" points at sine. If you have the two legs, "opposite + adjacent" points at tangent. Labeling the sides from the angle in the question is the single most important step — do it first, every time.

Finding a Missing Side

Suppose a right triangle has an acute angle of $35^\circ$ and a hypotenuse of $12$, and you want the side opposite that angle. You have the hypotenuse and want the opposite, so use sine:

$\sin 35^\circ = \frac{\text{opposite}}{12} \;\Rightarrow\; \text{opposite} = 12\sin 35^\circ.$

The SAT frequently leaves answers in this exact symbolic form, so recognizing which ratio to set up matters more than producing a decimal.

Finding a Missing Angle

If you know two sides and want the angle, use an inverse trig function. Given the opposite side $5$ and the adjacent side $12$, the tangent of the angle is $\tfrac{5}{12}$, so

$\theta = \tan^{-1}\!\left(\tfrac{5}{12}\right).$

Choose $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$ based on which two sides you were handed, using the same SOH-CAH-TOA logic in reverse.

The Sine–Cosine Cofunction Identity

The two acute angles of a right triangle add to $90^\circ$ — they are complementary. Because the side opposite one acute angle is adjacent to the other, the sine of one angle equals the cosine of the other:

$\sin\theta = \cos(90^\circ - \theta).$

This shows up as questions like "if $\sin(x^\circ) = 0.6$, what is $\cos(90 - x)^\circ$?" — and the answer is simply $0.6$, no triangle drawing needed. This identity is a favorite because it looks like it requires computation but rewards a student who recognizes the pattern.

A Fully Worked Example

A ramp rises at a $30^\circ$ angle to the ground, and its horizontal base measures $10$ feet. How high does the ramp reach? Give an exact expression.

\tan 30^\circ &= \frac{\text{height}}{10} \\ \text{height} &= 10\tan 30^\circ \\ &= 10\cdot \frac{\sqrt{3}}{3} = \frac{10\sqrt{3}}{3}\ \text{feet} \approx 5.77\ \text{feet.} \end{aligned}$$ The height is **opposite** the $30^\circ$ angle and the base is **adjacent** to it, so tangent is the right ratio. When the angle is $30^\circ$, $45^\circ$, or $60^\circ$, you can also switch to the [special right triangle](/test-prep/concepts/special-right-triangles) ratios and skip the trig function entirely, and every right triangle still obeys the [Pythagorean theorem](/test-prep/concepts/pythagorean-theorem) linking its three sides. ## Common Mistakes to Avoid - Labeling opposite and adjacent from the wrong angle. Always view the sides from the angle named in the question. - Reaching for sine when the sides you have call for cosine or tangent. - Forgetting that only the two acute angles are complementary — the right angle is not part of that pair. - Using the Pythagorean theorem when an angle is given and a single trig ratio is faster.