Trigonometric Ratios

Overview

Trigonometric ratios compare the sides of a right-angled triangle from the point of view of one acute angle. Once the angle is chosen, each side has a role: opposite, adjacent, or hypotenuse.

This topic matters because many distances and heights cannot be measured directly. Trigonometric ratios let a learner connect angles with lengths, prepare for Angles of Elevation and Depression, and later extend the same ideas to Sine Rule and Cosine Rule.

Prerequisites

• Ratios and proportions - Trigonometric ratios compare side lengths.
• Similarity and Congruence - Similar right triangles have equal angle measures and equal corresponding side ratios.
• Pythagoras' theorem - Missing sides in right triangles are often found before applying trigonometry.
• Angles - Learners should recognize acute angles, right angles, and degree notation such as $30^\circ$.
• Radicals - Exact special-angle values include surds such as $\sqrt{2}$ and $\sqrt{3}$.

Learning Scope

This chapter covers right-triangle trigonometric ratios, the meanings of $\sin \theta$, $\cos \theta$, and $\tan \theta$, reciprocal ratios, exact values for $30^\circ$, $45^\circ$, and $60^\circ$, and basic use of ratios to find missing sides and angles.

This page does not fully teach bearings, graphs of trigonometric functions, compound-angle identities, or the sine rule and cosine rule. Those belong to related later pages, although this chapter gives the foundation for them.

Subtopics

Naming Sides In A Right Triangle

In a right triangle, the hypotenuse is always the side opposite the right angle. It is the longest side.

The other two names depend on the chosen acute angle $\theta$:

• The opposite side is across from $\theta$.
• The adjacent side touches $\theta$ but is not the hypotenuse.

Key insight: The hypotenuse does not change when the angle changes, but opposite and adjacent can swap roles when a different acute angle is chosen.

Sine, Cosine, And Tangent

The three main trigonometric ratios are:

$$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} $$

$$ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} $$

$$ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} $$

A common memory aid is SOH-CAH-TOA:

• SOH: $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
• CAH: $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
• TOA: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

Key insight: A trigonometric ratio is not a side length by itself. It is a comparison of two side lengths.

Reciprocal Ratios

Some questions use the reciprocal ratios cosecant, secant, and cotangent:

$$ \csc \theta = \frac{1}{\sin \theta} = \frac{\text{hypotenuse}}{\text{opposite}} $$

$$ \sec \theta = \frac{1}{\cos \theta} = \frac{\text{hypotenuse}}{\text{adjacent}} $$

$$ \cot \theta = \frac{1}{\tan \theta} = \frac{\text{adjacent}}{\text{opposite}} $$

These are less common at the beginning of Form II work, but they are useful for recognizing equivalent expressions.

Choosing The Correct Ratio

To solve a right-triangle problem:

1. Mark the given angle.
2. Label the known and unknown sides as opposite, adjacent, or hypotenuse.
3. Choose the ratio that uses the two relevant sides.
4. Substitute values and solve.

For example, if a question gives an angle, the adjacent side, and asks for the opposite side, use tangent because:

$$ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} $$

Finding A Missing Side

If the angle and one side are known, a trigonometric ratio can find another side.

Suppose the adjacent side to angle $\theta$ is $12$ cm and the opposite side is $x$ cm. Then:

$$ \tan \theta = \frac{x}{12} $$

So:

$$ x = 12 \tan \theta $$

Key insight: Write the ratio first, then substitute. This prevents many wrong numerator and denominator choices.

Finding A Missing Angle

If two side lengths are known, the angle can be found using an inverse trigonometric function.

For example, if:

$$ \sin \theta = \frac{3}{5} $$

then:

$$ \theta = \sin^{-1}\left(\frac{3}{5}\right) $$

On a calculator, $\sin^{-1}$ gives the angle whose sine is the given ratio. In a right-triangle context, the acute angle is usually between $0^\circ$ and $90^\circ$.

Special-Angle Triangles

The exact values for $30^\circ$, $45^\circ$, and $60^\circ$ come from two standard right triangles.

From the $45^\circ$-$45^\circ$-$90^\circ$ triangle:

$$ \sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$

$$ \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$

$$ \tan 45^\circ = \frac{1}{1} = 1 $$

From the $30^\circ$-$60^\circ$-$90^\circ$ triangle:

$$ \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

$$ \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \tan 60^\circ = \sqrt{3} $$

Special-Angle Table

| Angle | $\sin \theta$ | $\cos \theta$ | $\tan \theta$ | | ---: | ---: | ---: | ---: | | $0^\circ$ | $0$ | $1$ | $0$ | | $30^\circ$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ | | $45^\circ$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ | | $60^\circ$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ | | $90^\circ$ | $1$ | $0$ | undefined |

The value of $\tan 90^\circ$ is undefined in right-triangle ratio form because:

$$ \tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0} $$

Division by zero is not defined.

Relationship Between The Main Ratios

For an acute angle in a right triangle:

$$ \tan \theta = \frac{\sin \theta}{\cos \theta} $$

This follows from the side definitions:

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

Also, because of Pythagoras' theorem:

$$ \sin^2 \theta + \cos^2 \theta = 1 $$

This identity is especially useful in later trigonometry, but the right-triangle meaning begins here.

Key Terms

• Trigonometric ratio: A ratio comparing two sides of a right triangle relative to a chosen acute angle.
• Hypotenuse: The longest side of a right triangle, opposite the right angle.
• Opposite side: The side across from the chosen acute angle.
• Adjacent side: The side next to the chosen acute angle, excluding the hypotenuse.
• Sine: The ratio $\frac{\text{opposite}}{\text{hypotenuse}}$.
• Cosine: The ratio $\frac{\text{adjacent}}{\text{hypotenuse}}$.
• Tangent: The ratio $\frac{\text{opposite}}{\text{adjacent}}$.
• Reciprocal ratio: A ratio formed by turning another ratio upside down, such as $\csc \theta = \frac{1}{\sin \theta}$.
• Exact value: A trigonometric value written without rounding, such as $\frac{\sqrt{3}}{2}$.
• Inverse trigonometric function: A function such as $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$ used to find an angle from a ratio.

Worked Examples

Example 1: Write The Three Main Ratios

A right triangle has angle $\theta$, opposite side $3$ cm, adjacent side $4$ cm, and hypotenuse $5$ cm. Find $\sin \theta$, $\cos \theta$, and $\tan \theta$.

Use the definitions:

$$ \begin{aligned} \sin \theta &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5} \\ \cos \theta &= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} \\ \tan \theta &= \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4} \end{aligned} $$

Final answer:

$$ \sin \theta = \frac{3}{5}, \quad \cos \theta = \frac{4}{5}, \quad \tan \theta = \frac{3}{4} $$

Example 2: Find A Missing Side Using Sine

In a right triangle, angle $A = 30^\circ$, the hypotenuse is $10$ cm, and the side opposite $A$ is $x$ cm. Find $x$.

Choose sine because opposite and hypotenuse are involved:

$$ \sin 30^\circ = \frac{x}{10} $$

Use $\sin 30^\circ = \frac{1}{2}$:

$$ \begin{aligned} \frac{1}{2} &= \frac{x}{10} \\ x &= 10 \times \frac{1}{2} \\ x &= 5 \end{aligned} $$

Final answer:

$$ x = 5 \text{ cm} $$

Example 3: Find A Missing Side Using Tangent

From a point on level ground, the angle to the top of a vertical pole is $60^\circ$. The point is $8$ m from the base of the pole. Find the height $h$ of the pole.

The height is opposite the angle and the ground distance is adjacent, so use tangent:

$$ \tan 60^\circ = \frac{h}{8} $$

Use $\tan 60^\circ = \sqrt{3}$:

$$ \begin{aligned} \sqrt{3} &= \frac{h}{8} \\ h &= 8\sqrt{3} \end{aligned} $$

Final answer:

$$ h = 8\sqrt{3} \text{ m} $$

Approximate interpretation:

$$ h \approx 13.9 \text{ m} $$

Example 4: Find An Angle From Two Sides

In a right triangle, the side opposite angle $\theta$ is $6$ cm and the hypotenuse is $10$ cm. Find $\theta$ to the nearest degree.

Use sine:

$$ \sin \theta = \frac{6}{10} = 0.6 $$

Apply the inverse sine:

$$ \theta = \sin^{-1}(0.6) $$

Using a calculator in degree mode:

$$ \theta \approx 36.9^\circ $$

Final answer:

$$ \theta \approx 37^\circ $$

Example 5: Use Pythagoras And Then Tangent

In triangle $ABC$, angle $B$ is a right angle, $CA = 6$ cm, and $AB = 3$ cm. Find $BC$ and $\tan C$.

Since $CA$ is opposite the right angle, it is the hypotenuse. Use Pythagoras' theorem:

$$ \begin{aligned} AB^2 + BC^2 &= CA^2 \\ 3^2 + BC^2 &= 6^2 \\ 9 + BC^2 &= 36 \\ BC^2 &= 27 \\ BC &= \sqrt{27} \\ BC &= 3\sqrt{3} \end{aligned} $$

For angle $C$, the opposite side is $AB = 3$ and the adjacent side is $BC = 3\sqrt{3}$:

$$ \begin{aligned} \tan C &= \frac{3}{3\sqrt{3}} \\ &= \frac{1}{\sqrt{3}} \\ &= \frac{\sqrt{3}}{3} \end{aligned} $$

Final answer:

$$ BC = 3\sqrt{3} \text{ cm}, \quad \tan C = \frac{\sqrt{3}}{3} $$

Example 6: Derive Tangent From Sine And Cosine

Show that $\frac{\sin \theta}{\cos \theta} = \tan \theta$ for an acute angle in a right triangle.

Start with the side-ratio definitions:

$$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} $$

Then divide:

$$ \begin{aligned} \frac{\sin \theta}{\cos \theta} &= \frac{\frac{\text{opposite}}{\text{hypotenuse}}}{\frac{\text{adjacent}}{\text{hypotenuse}}} \\ &= \frac{\text{opposite}}{\text{hypotenuse}} \times \frac{\text{hypotenuse}}{\text{adjacent}} \\ &= \frac{\text{opposite}}{\text{adjacent}} \\ &= \tan \theta \end{aligned} $$

Conclusion:

$$ \frac{\sin \theta}{\cos \theta} = \tan \theta $$

Common Mistakes

• Mistake: Labelling the opposite and adjacent sides before choosing the angle.

Correction: Choose the angle first; then label the sides relative to that angle.

• Mistake: Treating the hypotenuse as whichever side is drawn slanting upward.

Correction: The hypotenuse is opposite the right angle, no matter how the triangle is rotated.

• Mistake: Writing $\sin \theta = \frac{\text{hypotenuse}}{\text{opposite}}$.

Correction: $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$.

• Mistake: Using $\tan \theta$ when the known sides are opposite and hypotenuse.

Correction: Use sine for opposite and hypotenuse.

• Mistake: Rounding too early.

Correction: Keep exact values such as $\sqrt{3}$ or fractions until the final step unless the question asks for a decimal.

• Mistake: Using a calculator in radian mode when angles are in degrees.

Correction: For Form II degree-measure problems, set the calculator to degree mode.

• Mistake: Thinking $\sin^{-1} \theta$ means $\frac{1}{\sin \theta}$.

Correction: In calculator notation, $\sin^{-1}(x)$ means the inverse sine used to find an angle. The reciprocal of sine is $\csc \theta$.

• Mistake: Assuming $\tan 90^\circ = 1$ because $\sin 90^\circ = 1$.

Correction: $\tan 90^\circ$ is undefined because $\cos 90^\circ = 0$.

Practice Tasks

Direct Understanding

1. In a right triangle, which side is always the hypotenuse?
2. Define the opposite side relative to a chosen angle $\theta$.
3. Write the meanings of $\sin \theta$, $\cos \theta$, and $\tan \theta$.
4. Explain why opposite and adjacent can change when a different acute angle is chosen.

Skill Practice

1. A right triangle has opposite side $5$ cm, adjacent side $12$ cm, and hypotenuse $13$ cm relative to angle $\theta$. Find $\sin \theta$, $\cos \theta$, and $\tan \theta$.
2. Find $x$ if $\sin 30^\circ = \frac{x}{14}$.
3. Find $x$ if $\cos 60^\circ = \frac{x}{20}$.
4. Find $x$ if $\tan 45^\circ = \frac{x}{9}$.
5. Find $\theta$ to the nearest degree if $\tan \theta = \frac{4}{7}$.
6. Complete: $\csc \theta = \frac{1}{\sin \theta} = \frac{\text{?}}{\text{?}}$.

Application Problems

1. A ladder leans against a wall. The ladder is $10$ m long and makes a $60^\circ$ angle with the ground. Find the height reached on the wall.
2. A student stands $15$ m from the base of a tree. The angle to the top is $30^\circ$. Find the height of the tree in exact form.
3. A right triangle has hypotenuse $25$ cm and one leg $7$ cm. Find the other leg, then find the tangent of the angle opposite the $7$ cm side.
4. A ramp rises $1.5$ m over a horizontal distance of $6$ m. Find the angle the ramp makes with the ground to the nearest degree.
5. Without using a table, find $\frac{\sin 60^\circ}{\cos 60^\circ}$.

Multi-Step Reasoning

1. In a right triangle, $\cos A = \frac{5}{13}$ and $A$ is acute. Draw a possible triangle, find the missing side, and find $\tan A$.
2. If $\sin \theta = \frac{8}{17}$ for an acute angle, find $\cos \theta$ and $\tan \theta$.
3. Show using exact special-angle values that $\sin 30^\circ = \cos 60^\circ$.
4. Decide whether the statement $\sin \theta + \cos \theta = 1$ is always true for acute angles. Test it using $\theta = 45^\circ$.

Edge Cases

1. Why is $\tan 90^\circ$ undefined?
2. What changes if the angle is moved from one acute corner of a right triangle to the other?
3. A learner writes $\sin^{-1}(0.5) = 2$. Explain the error.
4. A diagram shows a side labelled $10$ cm across from the right angle. What is the name of that side?

Generated Question Layer

• Conceptual questions: Ask learners to label opposite, adjacent, and hypotenuse from different chosen angles.
• Ratio-identification questions: Generate tasks where learners choose sine, cosine, or tangent before substituting values.
• Exact-value questions: Use $30^\circ$, $45^\circ$, and $60^\circ$ with fractions and surds.
• Missing-side questions: Include direct right-triangle contexts and simple height or distance situations.
• Missing-angle questions: Use inverse trigonometric functions with degree-mode answers.
• Error-analysis questions: Include swapped numerator and denominator, wrong calculator mode, and confusion between inverse and reciprocal notation.
• Bridge questions: Combine Pythagoras' theorem with tangent or sine, preparing learners for angles of elevation and depression.

Learner Aid Opportunities

• diagram: Label opposite, adjacent, and hypotenuse from each acute angle in the same right triangle.
• chart: Summarize sine, cosine, tangent, and reciprocal ratios using SOH-CAH-TOA.
• animation: Show opposite and adjacent swapping when the chosen acute angle changes.
• interactive: Let learners choose a ratio after selecting the known side, unknown side, and angle.
• video: Demonstrate exact values from the $45^\circ$-$45^\circ$-$90^\circ$ and $30^\circ$-$60^\circ$-$90^\circ$ triangles.
• LLM tutor: Check whether a learner's ratio setup matches the side names and calculator mode.

Exam-Derived Signals

The automatic 2021-2025 mapping counted 4 primary topic-trigonometric-ratios records and 5 total trigonometry-hub records. These are unreviewed extraction signals, not verified official past-question links.

The 2022 examination format crosswalk maps the group Trigonometry and Pythagoras theorem to topic-trigonometric-ratios, topic-angles-of-elevation-and-depression, topic-sine-rule-and-cosine-rule, and topic-compound-angles, with Pythagoras theorem flagged as an unmatched term needing manual review.

Unreviewed extracted signals include:

| Year | Question ID | Signal | Review caution | | ---: | --- | --- | --- | | 2021 | csee_041_2021_p1_q09_a | Identity involving $\cos(90^\circ + \theta)$ and $\sin \theta$. | Secondary mapping only; also misclassified under exponents by the automatic mapper. | | 2022 | csee_041_2022_p1_q09_a_i | Evaluate $\frac{\sin 690^\circ}{\cos 690^\circ}$ without a table. | Secondary mapping only; beyond basic right-triangle ratios because it uses large-angle reduction. | | 2023 | csee_041_2023_p1_q09_b_i | Given $13\cos A - 5 = 0$ for acute $A$, find $\tan A$. | Stronger right-triangle ratio signal, but still unreviewed. | | 2024 | csee_041_2024_p1_q09_a | Show an exact value involving $\sin 75^\circ$. | Secondary mapping; belongs more naturally with compound-angle work than introductory ratios. | | 2025 | csee_041_2025_p1_q09_b | Right-triangle problem requiring a missing side and $\tan C$. | Stronger right-triangle ratio signal, but marked needs_manual_review and missing marks. |

Several automatic records near this topic appear to be false positives, including non-trigonometry accounting or linear-programming contexts. They should not be used as learner-facing past-question references until checked against the original papers.

Source And Review Notes

• Official syllabus status: The topic identity, Form II placement, competence, and hub come from data/curriculum_map.json, which cites raw/syllabuses/csee/2023/csee_mathematics_syllabus_2023.pdf.
• Official syllabus reference page: wiki/sources/csee-mathematics-syllabus-2023.md records the syllabus publisher, level, curriculum role, and assessment context.
• Exam format status: data/exam_format_topic_crosswalk_2022.jsonl is used as assessment guidance only. Its trigonometry grouping needs manual review because it includes an unmatched Pythagoras term.
• Exam signal status: data/question_map_2021_2025.jsonl and data/topic_frequency_2021_2025.json are unreviewed extraction outputs. The signals above are included for calibration, not as official solved past questions.
• Content status: The explanations, examples, and practice tasks on this page are original learner-facing prose created for this expansion.
• Renderer QA: This page uses portable $...$ and $$...$$ math notation for future Obsidian, KaTeX, or MathJax rendering.