# Sine, cosine, and tangent (right triangle) > AP Precalculus · Unit 3: Trigonometric and Polar Functions > Source: https://www.owlsprep.com/study/ap-precalculus-u3-sine-cosine-and-tangent/ This module covers the right triangle definition of sine, cosine, and tangent for AP Precalculus Unit 3. You will learn core ratios, solve for unknown sides and angles, apply cofunction identities, and avoid common exam traps. **Prerequisites:** Pythagorean theorem for right triangles; Basic angle measure in degrees and radians; Properties of similar triangles ## Learning objectives - Define sine, cosine, and tangent using right triangle side ratios - Solve for unknown sides and acute angles in right triangles - Apply cofunction identities for complementary angles - Recognize and avoid common exam pitfalls in right triangle trigonometry - Measure angles in radians as arc length over radius, and define tangent as the slope of the terminal ray in standard position ## Foundations: Right Triangles, Standard Position, and Radians Right triangle trigonometry provides the foundational definition of the three core trigonometric functions, based on constant side ratios of right-angled triangles. 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 triangle size. This definition applies to angles between $0^\circ$ and $90^\circ$ ($0$ and $\pi/2$ radians), before extending to all real numbers via the unit circle. Mastery here is required for all subsequent trigonometric work in AP Precalculus, making up 7-10% of total exam weight, appearing in both MCQ and FRQ sections. More generally, AP Precalculus defines these functions from an angle in **standard position** — its vertex at the origin, its initial ray along the positive $x$-axis, and a **terminal ray** produced by rotating the initial ray (counterclockwise for a positive angle, clockwise for a negative one). In this course angles are measured primarily in radians rather than degrees. **Radian Measure** — The radian measure of an angle equals the length of the arc it subtends divided by the radius of the circle: $\theta = \frac{s}{r}$, where $s$ is the arc length and $r$ is the radius. Because a full circle has circumference $2\pi r$, one full revolution is $2\pi$ radians and a straight angle is $\pi$ radians. *Example:* A central angle whose subtended arc length equals the radius measures exactly $1$ radian. When the terminal ray of an angle $\theta$ meets the unit circle at the point $(x, y)$, the cosine and sine of $\theta$ are exactly those coordinates, $\cos\theta = x$ and $\sin\theta = y$. The tangent of $\theta$ is then the **slope of the terminal ray**, equal to the ratio of these coordinates: $$\tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta} \quad (\text{slope of the terminal ray, } x \neq 0)$$ For an acute angle $0 < \theta < \frac{\pi}{2}$, these standard-position definitions agree exactly with the right-triangle ratios developed in the next section, so SOHCAHTOA is simply the special case of these more general definitions. ## Core Ratios and the SOHCAHTOA Definition In any right triangle, sides are labeled *relative to the acute angle $\theta$ you are analyzing*, not the right angle. The hypotenuse is always the side opposite the right angle (and the longest side). The opposite side is the leg that does not touch $\theta$, while the adjacent side is the leg that touches $\theta$. $$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{O}{H}$$ $$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{A}{H}$$ $$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{O}{A}$$ > **SOHCAHTOA Mnemonic** > > SOHCAHTOA (pronounced "so-ka-toe-ah") is the standard memory hook for these ratios: SOH = Sine = Opposite over Hypotenuse, CAH = Cosine = Adjacent over Hypotenuse, TOA = Tangent = Opposite over Adjacent. **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: $$AB^2 = AC^2 + BC^2 = 5^2 + 12^2 = 169 \implies AB = 13$$ 3. Calculate $\sin(A)$ using the core ratio: $$\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{12}{13}$$ 4. Calculate $\tan(A)$ using the core ratio: $$\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AC} = \frac{12}{5}$$ > **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. ## 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 SOHCAHTOA, the Pythagorean theorem, and the fact that the two acute angles sum to $90^\circ$. 1. To solve for an unknown side: (1) Identify known/needed sides labeled relative to the known angle, (2) Select the trig ratio connecting known and unknown side, (3) Set up and rearrange the equation to solve. 2. To solve for an unknown acute angle: (1) Identify the two known sides labeled 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. > **Exam Note** > > On the AP exam, always follow instructions for exact values vs. decimal approximations, and check that your answer makes sense (the hypotenuse must always be the longest side). **Worked example:** A right triangle has an acute angle of $32^\circ$, 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, which is cosine. 3. Substitute values into the ratio: $$\cos(32^\circ) = \frac{10}{c}$$ 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 $c \approx 11.8$ 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. *Calculator:* allowed ## Cofunction Identities for Right Triangles The two acute angles in a right triangle are complementary (they sum to $90^\circ$ or $\pi/2$ radians). For an acute angle $\theta$, the other acute angle is $90^\circ - \theta$, and the opposite and adjacent sides swap places relative to this new angle. This relationship 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 saving time on the exam. The name "cosine" is short for "complementary sine", referencing this relationship. **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 $90^\circ$ (or $\pi/2$ radians) in an expression, always check for a cofunction identity before reaching for your calculator to save time. ## AP-Style Worked Examples **Worked example:** 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 1. Label the triangle correctly: right angle at $Y$, so hypotenuse (opposite right angle) is $XZ = 10$. For target angle $Z$, opposite side is $XY$ and adjacent side is $YZ$. 2. We know $\tan(Z) = \frac{\text{opposite}}{\text{adjacent}} = \frac{XY}{YZ} = \frac{3}{4}$. Let $XY = 3k$ and $YZ = 4k$ for some positive constant $k$. 3. Apply the Pythagorean theorem: $$(3k)^2 + (4k)^2 = 10^2 \implies 25k^2 = 100 \implies k^2 = 4 \implies k = 2$$ 4. Substitute back to find $XY$: $$XY = 3(2) = 6$$ **Worked example:** 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 $\tan(Q)$. 1. Part (a): The sum of interior angles in any triangle is $180^\circ$, so: $$90^\circ + 2\theta + (3\theta - 10^\circ) = 180^\circ$$ 2. Simplify and solve for $\theta$: $$5\theta + 80^\circ = 180^\circ \implies 5\theta = 100^\circ \implies \theta = 20^\circ$$ 3. The acute angles are $2\theta = 40^\circ$ and $3(20^\circ) - 10^\circ = 50^\circ$. 4. Part (b): $PR$ is adjacent to angle $P = 40^\circ$, so use the cosine ratio: $$\cos(P) = \frac{PR}{PQ} \implies PQ = \frac{PR}{\cos(P)} = \frac{7}{\cos(40^\circ)}$$ 5. This is the required exact value ($7\sec(40^\circ)$ is also acceptable). 6. Part (c): $Q = 90^\circ - P = 50^\circ$, so apply the cofunction identity: $$\tan(Q) = \tan(90^\circ - 40^\circ) = \cot(40^\circ) = \frac{1}{\tan(40^\circ)}$$ 7. This is the required exact value. **Worked example:** Real-World Application: 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 $12.5^\circ$. 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? 1. This scenario forms a right triangle where the $12.5^\circ$ 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 $h$. Tangent relates opposite and adjacent sides: $$\tan(12.5^\circ) = \frac{h}{1500}$$ 2. Rearrange to solve for $h$: $$h = 1500 \tan(12.5^\circ)$$ 3. Calculate with a calculator in degree mode: $\tan(12.5^\circ) \approx 0.2217$, so $h \approx 1500(0.2217) = 332.55$ m. 4. Add the height of the measuring device: $$\text{Total height} \approx 332.55 + 1.8 = 334.35 \approx 334 \text{ m}$$ 5. The total height of the peak above the plain is approximately 334 meters. *Calculator:* allowed ## Common pitfalls - **Wrong:** Labeling opposite/adjacent sides relative to the right angle instead of the target acute angle - Why it fails: Students rush to label the "bottom side" as adjacent without confirming which angle they are solving for. - Correct: Every time you start a problem, explicitly mark your target $\theta$, then label opposite/adjacent only relative to that angle. - **Wrong:** 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 it fails: Students forget to rearrange the equation correctly after setting up the ratio. - Correct: After solving for any side, check that the hypotenuse is longer than either leg; if not, reverse your ratio. - **Wrong:** Leaving the calculator in radian mode when the problem uses degrees, or vice versa - Why it fails: Students keep the mode from a previous problem and forget to switch. - Correct: Before any trig calculation, explicitly check your calculator's mode matches the angle unit given in the problem. - **Wrong:** Using the right triangle definition for angles greater than or equal to 90° - Why it fails: Students memorize SOHCAHTOA first and forget its domain restriction. - Correct: For angles outside $0 < \theta < 90^\circ$, use the unit circle definition instead of right triangle ratios. - **Wrong:** Picking the wrong hypotenuse for non-standard triangle orientations, assuming it is always horizontal - Why it fails: Students expect the hypotenuse to always be the horizontal or bottom side, not the side opposite the right angle. - Correct: Always confirm the right angle first, then the hypotenuse is always opposite the right angle, regardless of how the triangle is drawn. ## Cheatsheet | Category | Formula | Notes | | --- | --- | --- | | Sine ratio (right triangle) | $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$ | Only for $0 < \theta < 90^\circ$ ($0 < \theta < \frac{\pi}{2}$ radians); all values positive | | Cosine ratio (right triangle) | $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ | Only for $0 < \theta < 90^\circ$ ($0 < \theta < \frac{\pi}{2}$ radians); all values positive | | Tangent ratio (right triangle) | $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ | Undefined at $\theta = 90^\circ$; only for $0 < \theta < 90^\circ$ | | SOHCAHTOA Mnemonic | Sine=Opposite/Hypotenuse, Cosine=Adjacent/Hypotenuse, Tangent=Opposite/Adjacent | Use to recall core ratio definitions | | Cofunction Identities | $\sin\theta = \cos(90^\circ - \theta), \cos\theta = \sin(90^\circ - \theta)$ | Holds for any acute $\theta$ in a right triangle | | Solving for unknown angle | If $\sin\theta = k$, $\theta = \arcsin(k)$; if $\cos\theta = k$, $\theta = \arccos(k)$; if $\tan\theta = k$, $\theta = \arctan(k)$ | Use inverse trig to find angles from known side ratios | | Complementary angle sum | $\theta_1 + \theta_2 = 90^\circ$ | For the two acute angles in any right triangle | | Radian measure | $\theta = \frac{s}{r}$ | Arc length over radius; one full revolution $= 2\pi$ radians | | Tangent as slope (standard position) | $\tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta}$ | Slope of the terminal ray through $(x,y)$; undefined when $x = 0$ | ## 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 $90^\circ$, 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 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. - [Inverse trigonometric functions](https://www.owlsprep.com/study/ap-precalculus-u3-inverse-trigonometric-functions/) - [Sine and cosine function values (unit circle)](https://www.owlsprep.com/study/ap-precalculus-u3-sine-and-cosine-function-values/) - [Sine and cosine function graphs](https://www.owlsprep.com/study/ap-precalculus-u3-sine-and-cosine-function-graphs/) --- From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ap-precalculus-u3-sine-cosine-and-tangent/