Sine, cosine, and tangent (right triangle) — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Right triangle definition of sine, cosine, and tangent; SOHCAHTOA labeling and mnemonic; solving for unknown sides and angles; cofunction identities; applied right triangle trigonometry problems for Unit 3.

You should already know: Pythagorean theorem for right triangles, basic angle measure in degrees and radians, properties of similar triangles.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Sine, cosine, and tangent (right triangle)?

Right triangle trigonometry provides the first, foundational definition of the three core trigonometric functions, based on constant side ratios of a right-angled triangle (one interior angle equal to 90° or π/2 radians). Per the AP Precalculus Course and Exam Description (CED), this topic is part of Unit 3: Trigonometric and Polar Functions, making up roughly 7-10% of total exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. The core idea relies on similarity: all right triangles with a congruent acute angle are similar, so the ratio of any two sides is constant for a given angle, regardless of the triangle’s overall size. This constant ratio defines sine, cosine, and tangent for angles between 0° and 90° (0 and π/2 radians), before we extend these functions to all real numbers via the unit circle. Standard AP notation is $\sin \theta$ for sine of $\theta$, $\cos \theta$ for cosine, and $\tan \theta$ for tangent, with no alternative notation expected or accepted on the exam. Mastery of this topic is required for almost all subsequent trigonometric work in the course.

2. Core Ratios and the SOHCAHTOA Definition

In any right triangle, we label sides relative to the acute angle $\theta$ we are analyzing, not relative to the right angle. The hypotenuse is always the side opposite the right angle, and it is always the longest side of the triangle. The opposite side is the leg that does not touch $\theta$, while the adjacent side is the leg that does touch $\theta (it sits next to the target angle).

With this labeling, the three core trigonometric ratios are defined as: $$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{O}{H}$$ $$\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{A}{H}$$ $$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{O}{A}$$

The mnemonic SOHCAHTOA (pronounced "so-ka-toe-ah") is the standard way to remember these definitions: Sine = Opposite over Hypotenuse, Cosine = Adjacent over Hypotenuse, Tangent = Opposite over Adjacent. This definition only applies to acute angles between 0° and 90° in a right triangle, and all ratios are positive for this domain, since all side lengths are positive.

Worked Example

In right triangle $ABC$ with a right angle at $C$, $AC=5$ and $BC=12$. Find $\sin (A)$ and $\tan (A)$.

1. Label sides relative to target angle $A$: The right angle is at $C$, so hypotenuse is $AB$. The side opposite $A$ is $BC$, and the side adjacent to $A$ is $AC$.
2. Calculate hypotenuse length with the Pythagorean theorem: $A{B}^2=A{C}^2+B{C}^2=5^2+12^2=169$, so $AB=13$.
3. Calculate $\sin (A)$: $\sin (A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{12}{13}$.
4. Calculate $\tan (A)$: $\tan (A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AC}=\frac{12}{5}$.

Exam tip: Always label your sides relative to the target angle, not the other acute angle. Mixing up opposite and adjacent is the most common error on basic ratio problems, and it is a common MCQ distractor.

3. Solving for Unknown Sides and Angles

Once you know at least one side and one acute angle of a right triangle, you can find all unknown sides and angles using a combination of SOHCAHTOA, the Pythagorean theorem, and the fact that the two acute angles in a right triangle sum to 90°.

To solve for an unknown side:

1. Identify which sides you know, which side you need, and label them relative to the known angle.
2. Select the trig ratio that connects the known and unknown side.
3. Set up the equation and rearrange to solve for the unknown side.

To solve for an unknown acute angle:

1. Identify the two known sides relative to the unknown angle.
2. Set up the matching trig ratio, then use the corresponding inverse trigonometric function ($\arcsin$, $\arccos$, $\arctan$) to solve for the angle. Inverse trig functions "undo" the trig ratio: if $\sin \theta =k$, then $\theta =\arcsin (k)$ for acute $\theta$.

On the AP exam, always follow instructions for exact values vs. decimal approximations, and check that your answer makes physical sense (the hypotenuse must always be the longest side).

Worked Example

A right triangle has an acute angle of 32°, and the side adjacent to this angle is 10 cm. Find the length of the hypotenuse, to one decimal place.

1. Label knowns and unknowns: $\theta =32^{\circ }$, known adjacent side = 10 cm, unknown hypotenuse = $c$.
2. Select the trig ratio that connects adjacent and hypotenuse: this is cosine, $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$.
3. Substitute values: $\cos (32^{\circ })=\frac{10}{c}$.
4. Rearrange to solve for $c$: $c=\frac{10}{\cos (32^{\circ })}$.
5. Calculate with a calculator in degree mode: $\cos (32^{\circ })\approx 0.8480$, so $c\approx \frac{10}{0.8480}\approx 11.8$ cm.

Exam tip: Always check your calculator's angle mode before calculating trig or inverse trig values; mixing degrees and radians will always give an incorrect answer that matches one of the MCQ distractors.

4. Cofunction Identities for Right Triangles

Because the two acute angles in a right triangle are complementary (they sum to 90° or π/2 radians), we can derive a set of simple identities relating the trig functions of an angle and its complement. If we have one acute angle $\theta$, the other is $90^{\circ }-\theta$. For this second angle, the side that was opposite $\theta$ becomes the adjacent side, and vice versa.

This gives us the cofunction identities: $$\sin (\theta )=\cos (90^{\circ }-\theta )$$ $$\cos (\theta )=\sin (90^{\circ }-\theta )$$ $$\tan (\theta )=\cot (90^{\circ }-\theta )$$

These identities are useful for simplifying trigonometric expressions, connecting complementary angles, and verifying other trigonometric rules. They also explain the origin of the name "cosine": it is short for "complementary sine".

Worked Example

Simplify $\sin (47^{\circ })-\cos (43^{\circ })$ without using a calculator.

1. Check the sum of the angles: $47^{\circ }+43^{\circ }=90^{\circ }$, so $43^{\circ }=90^{\circ }-47^{\circ }$.
2. Apply the cofunction identity for cosine: $\cos (90^{\circ }-\theta )=\sin (\theta )$, so $\cos (43^{\circ })=\cos (90^{\circ }-47^{\circ })=\sin (47^{\circ })$.
3. Substitute back into the original expression: $\sin (47^{\circ })-\sin (47^{\circ })=0$.
4. The simplified value of the expression is 0.

Exam tip: If you see two angles that add up to 90° (or π/2 radians) in an expression, always check for a cofunction identity before reaching for your calculator to save time on the exam.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Labeling opposite/adjacent sides relative to the right angle instead of the target acute angle. Why: Students rush to label the "bottom side" as adjacent without confirming which angle they are solving for. Correct move: Every time you start a problem, explicitly mark your target $\theta$, then label opposite/adjacent only relative to that angle.
• Wrong move: Flipping the ratio when solving for an unknown side, e.g. writing $c=10\cos (32^{\circ })$ instead of $c=\frac{10}{\cos (32^{\circ })}$. Why: Students forget to rearrange the equation correctly after setting up the ratio. Correct move: After solving for any side, check that the hypotenuse is longer than either leg; if not, reverse your ratio.
• Wrong move: Leaving the calculator in radian mode when the problem uses degrees, or vice versa. Why: Students keep the mode from a previous problem and forget to switch. Correct move: Before any trig calculation, explicitly check your calculator's mode matches the angle unit given in the problem.
• Wrong move: Using the right triangle definition for angles greater than or equal to 90°. Why: Students memorize SOHCAHTOA first and forget its domain restriction. Correct move: For angles outside $0<\theta <90^{\circ }$, use the unit circle definition instead of right triangle ratios.
• Wrong move: Assuming the longest side is opposite the largest angle in a right triangle, and picking the wrong hypotenuse for non-standard triangle orientations. Why: Students expect the hypotenuse to always be the horizontal or bottom side, not the side opposite the right angle. Correct move: Always confirm the right angle first, then the hypotenuse is always opposite the right angle, regardless of how the triangle is drawn.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

In right triangle $XYZ$ with right angle at $Y$, $XZ=10$, and $\tan (Z)=\frac{3}{4}$. What is the length of $XY$? A) 4 B) 6 C) 8 D) 10

Worked Solution: First, label the triangle correctly: the right angle is at $Y$, so the hypotenuse (opposite the right angle) is $XZ=10$, which matches the problem statement. For target angle $Z$, the opposite side is $XY$ and the adjacent side is $YZ$. We know $\tan (Z)=\frac{\text{opposite}}{\text{adjacent}}=\frac{XY}{YZ}=\frac{3}{4}$, so we can write $XY=3k$ and $YZ=4k$ for some positive constant $k$. By the Pythagorean theorem: $(3k{)}^2+(4k{)}^2=10^2\to 25k^2=100\to k^2=4\to k=2$. Substituting back, $XY=3(2)=6$. The correct answer is B.

Question 2 (Free Response)

Right triangle $PQR$ has a right angle at $R$. The measure of angle $P$ is $2\theta$, and the measure of angle $Q$ is $(3\theta -10{)}^{\circ }$, with all angles in degrees. (a) Show that $\theta =20$, and find the measures of the two acute angles. (b) If $PR=7$, find the exact length of hypotenuse $PQ$. (c) Find the exact value of $\tan (Q)$.

Worked Solution: (a) The sum of interior angles in any triangle is $180^{\circ }$, so: $$90^{\circ }+2\theta +(3\theta -10^{\circ })=180^{\circ }$$ Simplify: $5\theta +80^{\circ }=180^{\circ }\to 5\theta =100^{\circ }\to \theta =20^{\circ }$, as required. The acute angles are $2\theta =40^{\circ }$ and $3(20^{\circ })-10^{\circ }=50^{\circ }$. (b) $PR$ is the side adjacent to angle $P=40^{\circ }$, so $\cos (P)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{PR}{PQ}$. Rearranging gives $PQ=\frac{PR}{\cos (P)}=\frac{7}{\cos (40^{\circ })}$ (or $7\sec (40^{\circ })$, both are acceptable exact values). (c) Using the cofunction identity: $Q=90^{\circ }-P=90^{\circ }-40^{\circ }=50^{\circ }$, so $\tan (Q)=\tan (90^{\circ }-40^{\circ })=\cot (40^{\circ })=\frac{1}{\tan (40^{\circ })}$, which is the exact value.

Question 3 (Application / Real-World Style)

A surveyor needs to find the height of a mountain peak above a flat coastal plain. They place a laser measuring device 1500 meters horizontally from the point directly below the peak, and measure the angle of elevation from the device to the peak to be 12.5 degrees. The measuring device is 1.8 meters above the plain. What is the total height of the peak above the plain, to the nearest meter?

Worked Solution: This scenario forms a right triangle where the 12.5° angle of elevation has an adjacent side equal to the horizontal distance of 1500 m, and the opposite side is the height of the peak above the measuring device, called $h$. We use tangent, which relates opposite and adjacent sides: $$\tan (12.5^{\circ })=\frac{h}{1500}$$ Rearrange to solve for $h$: $h=1500\tan (12.5^{\circ })$. With a calculator in degree mode, $\tan (12.5^{\circ })\approx 0.2217$, so $h\approx 1500(0.2217)=332.55$ m. Add the height of the measuring device: total height $\approx 332.55+1.8=334.35$ m, which rounds to 334 m. In context, the mountain peak is approximately 334 meters above the surrounding coastal plain.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational base for all further trigonometry in AP Precalculus. Next you will extend the right triangle definitions of sine, cosine, and tangent to all real angles using the unit circle, which allows you to work with angles larger than 90° and negative angles, and use trigonometric functions to model periodic phenomena. Without mastering right triangle SOHCAHTOA, you will struggle to connect unit circle coordinates to trig values and to solve applied trig problems that require side/angle calculations. This topic also feeds directly into the Laws of Sines and Cosines, which let you solve non-right triangles, and into polar coordinates, where sine and cosine are used to convert between polar and rectangular form.

Unit circle definition of trigonometric functions Inverse trigonometric functions Law of Sines and Cosines Polar coordinate conversions