Trigonometry - ACT Math | Varsity Tutors

SECTION 1

Historical Context & Motivation

Long before anyone coined the word "trigonometry," ancient civilizations needed a reliable way to connect angles to measurable distances. Egyptian builders calculated the slopes of pyramid faces, Babylonian astronomers tracked planetary motion across the night sky, and Greek navigators charted courses across the Mediterranean. Each group discovered the same core idea independently: if you know a triangle's angles, you can predict its side lengths, and vice versa. That insight is the heartbeat of trigonometry, and it remains one of the most tested topics on the ACT Math section.

~1800 BCE • Babylonian Angle Tables

Babylonian scribes carved clay tablets listing the ratios of triangle sides, creating some of the earliest known trigonometric-style tables.

~150 BCE • Hipparchus & the Chord Table

Greek astronomer Hipparchus compiled a table of chord lengths in a circle, laying the groundwork for the sine function and earning the title 'father of trigonometry.'

~500 CE • Indian Sine Function

Indian mathematician Aryabhata introduced the 'half-chord' concept, which evolved into the modern sine function used throughout trigonometry today.

~1400 CE • Islamic Golden Age Advances

Scholars such as al-Kashi computed sine tables to remarkable precision and developed all six trigonometric ratios, bridging ancient geometry and modern math.

1595 • Modern Notation Emerges

Bartholomeo Pitiscus published 'Trigonometria,' coining the term we use today and formalizing the subject as its own branch of mathematics.

The central question that drove all of this work is deceptively simple: Given a triangle with certain known measurements, how can we find the unknown sides and angles? The ACT tests your ability to answer exactly this question using the trigonometric ratios and identities we will explore in the sections ahead.

SECTION 2

Core Principles & Definitions

Trigonometry is built on the idea that a right triangle (a triangle with one 90° angle) has fixed ratios between its sides for any given acute angle. Once you memorize the three primary ratios—sine, cosine, and tangent—you can solve almost every trig problem the ACT throws at you. These ratios are collectively remembered with the mnemonic SOH-CAH-TOA, and they form the foundation for everything else in this lesson.

Sine (sin θ)

The ratio of the side opposite the angle to the hypotenuse. SOH: Sin = Opposite / Hypotenuse.

Cosine (cos θ)

The ratio of the side adjacent to the angle to the hypotenuse. CAH: Cos = Adjacent / Hypotenuse.

Tangent (tan θ)

The ratio of the side opposite the angle to the side adjacent. TOA: Tan = Opposite / Adjacent.

The Hypotenuse

Always the longest side of a right triangle, located directly across from the 90° angle. It serves as the denominator in both the sine and cosine ratios.

Complementary Relationship

In a right triangle the two acute angles add to 90°. The sine of one angle always equals the cosine of the other: sin θ = cos(90° − θ).

✦

KEY TAKEAWAY

Think of SOH-CAH-TOA like a recipe card. Just as a recipe tells you the ratio of flour to sugar to make a cake, SOH-CAH-TOA tells you the ratio of specific triangle sides for a given angle. Change the angle and the 'recipe' changes—but the format stays the same, which is exactly what makes it so powerful on test day.

SECTION 3

Visual Explanation — The Right Triangle

In this right triangle, angle θ is at vertex A. The side directly across from θ (side BC) is the opposite. The side touching θ that is not the hypotenuse (side AB) is the adjacent. The longest side across from the 90° angle (side AC) is the hypotenuse.

Notice that the labels opposite and adjacent depend entirely on which acute angle you are considering. If you shift your attention to the other acute angle in the triangle, the opposite and adjacent sides swap. The hypotenuse, however, never changes because it is always the side across from the right angle. This distinction is the number-one source of careless errors on the ACT, so practice identifying the correct sides before you plug anything into a formula.

SECTION 4

Mathematical Framework

The ACT expects you to know the three primary trig ratios, the Pythagorean identity, and how to convert between degrees and radians. Below are the key equations you should commit to memory, along with explanations of each variable.

PRIMARY TRIGONOMETRIC RATIOS

sin θ = opposite / hypotenuse cos θ = adjacent / hypotenuse tan θ = opposite / adjacent

θ (theta) represents the acute angle of interest. Opposite is the side across from θ, adjacent is the side next to θ (not the hypotenuse), and hypotenuse is the longest side.

PYTHAGOREAN IDENTITY

sin²θ + cos²θ = 1

This identity is derived directly from the Pythagorean theorem (a² + b² = c²) by dividing every term by the hypotenuse squared. It lets you find sin θ when you know cos θ, and vice versa.

TANGENT IN TERMS OF SINE AND COSINE

tan θ = sin θ / cos θ

Because tangent is opposite over adjacent, and sine over cosine equals (opp/hyp) ÷ (adj/hyp), the hypotenuse cancels. This relationship is useful when the ACT gives you sin θ and cos θ and asks for tan θ.

DEGREE-RADIAN CONVERSION

radians = degrees × (π / 180) degrees = radians × (180 / π)

A full circle is 360° or 2π radians. Common ACT conversions: 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2.

💡

ACT Tip

Most ACT trig problems use the special right triangles: 30-60-90 (side ratios 1 : √3 : 2) and 45-45-90 (side ratios 1 : 1 : √2). Memorizing these side ratios means you can write down sin, cos, and tan of 30°, 45°, and 60° instantly—no calculator required.

SECTION 5

The Unit Circle & Special Angles

While SOH-CAH-TOA works perfectly for right triangles, the unit circle extends trigonometry to angles of any size—including those greater than 90° and negative angles. The unit circle is a circle of radius 1 centered at the origin of the coordinate plane. Every point on the circle can be written as (cos θ, sin θ), where θ is the angle measured counterclockwise from the positive x-axis. The ACT sometimes asks questions about trig values in different quadrants, so understanding how signs change around the circle is essential.

The unit circle shows that every point is (cos θ, sin θ). In Quadrant I both sin and cos are positive. In Quadrant II only sin is positive. In Quadrant III both are negative. In Quadrant IV only cos is positive. A common mnemonic is "All Students Take Calculus" (All positive, Sine positive, Tangent positive, Cosine positive).

Angle (°)	Angle (rad)	sin θ	cos θ	tan θ
0°	0	0	1	0
30°	π/6	1/2	√3/2	√3/3
45°	π/4	√2/2	√2/2	1
60°	π/3	√3/2	1/2	√3
90°	π/2	1	0	undefined

Trig values for the most commonly tested angles on the ACT

SECTION 6

Worked Example

Let's walk through a typical ACT-style problem step by step. Suppose you're given a right triangle where one acute angle is 35° and the hypotenuse measures 20 cm. The question asks you to find the length of the side opposite the 35° angle.

Finding a Missing Side Using Sine

1
Step 1 — Identify the Given Information

You know the angle θ = 35°, and the hypotenuse = 20 cm. You need the side opposite the 35° angle.
2
Step 2 — Choose the Right Ratio

You have the hypotenuse and need the opposite side. The ratio that connects opposite and hypotenuse is sine: sin θ = opposite / hypotenuse.
3
Step 3 — Substitute Known Values

Plug in what you know: sin 35° = opposite / 20.
4
Step 4 — Solve for the Unknown

Multiply both sides by 20: opposite = 20 × sin 35°. Using a calculator, sin 35° ≈ 0.5736, so opposite = 20 × 0.5736.

opposite ≈ 11.47 cm
5
Step 5 — Verify Reasonableness

The opposite side (≈ 11.47 cm) is shorter than the hypotenuse (20 cm), which makes sense because no leg of a right triangle can be longer than the hypotenuse. Additionally, 35° is less than 45°, so the opposite side should be less than half of 20 × √2 ≈ 14.14, and 11.47 fits that expectation.

⚠

Common Mistake Alert

Many students accidentally use cosine when they need sine, or vice versa. Before you calculate, always circle the two triangle parts you're working with (one known, one unknown) and pick the ratio that connects them. If you have the adjacent and need the hypotenuse, that's cosine—not sine.

SECTION 7

Strengths, Limitations & ACT Strategies

SOH-CAH-TOA is incredibly effective for right triangles, but trigonometry on the ACT goes a bit further. Understanding where each tool shines—and where it falls short—will help you choose the fastest approach on test day.

Tool / Concept	Strengths	Limitations
SOH-CAH-TOA	Fast and direct for any right triangle. Only requires one angle and one side to find another side.	Only applies to right triangles. Cannot be used directly for obtuse or non-right triangles.
Unit Circle	Handles any angle (including > 90°, negatives, and radians). Essential for determining signs in different quadrants.	Requires memorization of key coordinates. Does not directly give side lengths of a specific triangle.
Pythagorean Identity	Lets you find sin from cos (or vice versa) without a calculator. Useful when the ACT gives you one ratio.	Only connects sin and cos. You still need the tangent identity or other information to find tan.
Law of Sines / Law of Cosines	Extends trig to any triangle, not just right triangles. Occasionally tested on the ACT.	More complex setup. Requires careful attention to which sides and angles correspond.

🎯

ACT STRATEGY

Think of your trig toolkit like a GPS app. SOH-CAH-TOA is the quickest route for right-triangle problems—it gets you there in the fewest steps. The unit circle is your satellite view, giving you the big picture of any angle. The Pythagorean identity is a shortcut that avoids recalculating from scratch. On the ACT, identify which tool matches the problem before you start calculating; choosing the right tool first saves more time than calculating faster.

SECTION 8

Connection to Advanced Trigonometry

The ACT occasionally dips into topics that go beyond basic right-triangle trig. The table below shows how the foundational concepts you've learned connect to these more advanced ideas. You don't need to master all of them, but recognizing these connections can help you handle the hardest questions on the test.

Foundational Concept	Advanced Extension	When It Appears on the ACT
SOH-CAH-TOA (basic ratios)	Reciprocal functions: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ	Rarely tested; may appear in the hardest 5 questions.
Unit circle (sin and cos of standard angles)	Graphing y = sin x and y = cos x; amplitude, period, and phase shifts	1–2 questions per test on recognizing or interpreting trig graphs.
sin²θ + cos²θ = 1	Double-angle identities: sin 2θ = 2 sin θ cos θ, cos 2θ = cos²θ − sin²θ	Occasionally tested; may appear as simplification or substitution problems.
Right-triangle applications	Law of Sines: a/sin A = b/sin B; Law of Cosines: c² = a² + b² − 2ab cos C	Tested 0–1 times per exam, but can be a scoring differentiator.

If you're aiming for a score in the 30+ range, spend some time familiarizing yourself with trig graphs and the Law of Sines and Cosines. These topics bridge the gap between geometry-level trig and the precalculus material you'll encounter in college. Even if these advanced questions appear only once or twice, they are often worth the same number of points as easier questions, and mastering them can set you apart.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

In a right triangle, angle A is an acute angle. If sin A = 3/5, what is cos A? Explain your reasoning without using a calculator.

Show answer

Using the Pythagorean identity, sin²A + cos²A = 1. Substituting sin A = 3/5, you get (3/5)² + cos²A = 1, which simplifies to 9/25 + cos²A = 1. Solving gives cos²A = 16/25, so cos A = 4/5 (positive because A is acute and therefore in Quadrant I). Alternatively, you could recognize the 3-4-5 right triangle: the opposite side is 3, the hypotenuse is 5, and the adjacent side must be 4, giving cos A = 4/5.

Question 2: BASIC CALCULATION

A right triangle has an acute angle of 40° and the side adjacent to that angle is 15 cm long. What is the length of the hypotenuse? Round to the nearest tenth.

Show answer

You know the adjacent side and need the hypotenuse, so use cosine: cos 40° = adjacent / hypotenuse = 15 / hypotenuse. Solving for the hypotenuse: hypotenuse = 15 / cos 40°. Since cos 40° ≈ 0.7660, the hypotenuse = 15 / 0.7660 ≈ 19.6 cm. This makes sense because the hypotenuse must be longer than the adjacent side, and 19.6 > 15.

Question 3: INTERMEDIATE

If tan θ = 7/24 and θ is in Quadrant I, find the exact values of sin θ and cos θ.

Show answer

Since tan θ = opposite/adjacent = 7/24, think of a right triangle with opposite = 7 and adjacent = 24. Use the Pythagorean theorem to find the hypotenuse: h = √(7² + 24²) = √(49 + 576) = √625 = 25. Therefore, sin θ = 7/25 and cos θ = 24/25. You can verify: tan θ = sin θ / cos θ = (7/25) / (24/25) = 7/24 ✓, and sin²θ + cos²θ = 49/625 + 576/625 = 625/625 = 1 ✓.

Question 4: APPLIED

A surveyor stands 80 meters from the base of a building and measures the angle of elevation to the top of the building to be 54°. How tall is the building? Round to the nearest meter.

Show answer

Draw a right triangle where the horizontal distance (80 m) is the side adjacent to the 54° angle and the building height is the side opposite the angle. Use tangent: tan 54° = height / 80. Therefore, height = 80 × tan 54° ≈ 80 × 1.3764 ≈ 110 meters. This is a classic angle-of-elevation problem that appears frequently on the ACT. Always check that your answer is reasonable: a 54° angle is steeper than 45°, so the building should be taller than 80 m, which checks out.

Question 5: CRITICAL THINKING

Angle θ is in Quadrant II and sin θ = 5/13. Find the exact value of tan θ. Explain why the sign of your answer makes sense in the context of the unit circle.

Show answer

Start with the Pythagorean identity: sin²θ + cos²θ = 1, so (5/13)² + cos²θ = 1, giving cos²θ = 1 − 25/169 = 144/169. Thus cos θ = ±12/13. Since θ is in Quadrant II, cosine is negative, so cos θ = −12/13. Now use tan θ = sin θ / cos θ = (5/13) / (−12/13) = −5/12. The negative sign makes sense because in Quadrant II, sine is positive and cosine is negative, and a positive divided by a negative yields a negative result. This aligns with the 'All Students Take Calculus' mnemonic: only sine is positive in QII, so tangent must be negative there.

SUMMARY

Lesson Summary

Summary

Trigonometry connects angles to side lengths through three fundamental ratios: sine (opposite / hypotenuse), cosine (adjacent / hypotenuse), and tangent (opposite / adjacent). The mnemonic SOH-CAH-TOA is your go-to tool for right-triangle problems, while the unit circle extends these ratios to angles of any size by defining every point on a circle of radius 1 as (cos θ, sin θ).

Key formulas to remember include the Pythagorean identity (sin²θ + cos²θ = 1) and the relationship tan θ = sin θ / cos θ. Memorize the trig values of special angles (30°, 45°, 60°) and know the sign rules for each quadrant using "All Students Take Calculus." On the ACT, always identify which sides and angles you have before choosing a ratio, and double-check that your answer is reasonable by confirming that no leg exceeds the hypotenuse.