---
title: Right-Triangle Trigonometry
subtitle: Proofs, exact values, solved triangles, and applications
author: Mathematics Department
organization: Your School
description: A proof-led lesson on right-triangle ratios, exact values at 30, 45, and 60 degrees, identities, inverse trigonometry, and heights and distances.
keywords: [right triangle trigonometry, sine, cosine, tangent, exact values, special triangles, angle of elevation, angle of depression]
template: assignment
theme: latex-article
paper: us-letter
titlePage: true
toc: true
headingNumbers: false
equationNumbers: false
columns: 1
figurePlacement: none
tableCaptionPosition: top
footer: Student lesson and practice
pdfProfile: tagged
language: en-US
---

# Right-triangle trigonometry

Trigonometry connects the shape of a right triangle with numerical ratios. The ratios are not arbitrary rules: they are forced by similarity. Once one acute angle is fixed, every right triangle containing that angle has the same shape, so its corresponding side ratios are constant.

## Sides relative to an angle

In a right triangle, the **hypotenuse** is opposite the right angle and is always the longest side. Relative to a chosen acute angle $\theta$:

- the **opposite** side lies across from $\theta$;
- the **adjacent** side touches $\theta$ but is not the hypotenuse.

In @fig:ratios, write the opposite, adjacent, and hypotenuse lengths as $y$, $x$, and $h$.

:::figure{#fig:ratios caption="The side names and two similar right triangles containing the same acute angle" alt="A right triangle labels the side across from theta as y, the horizontal side next to theta as x, and the hypotenuse as h; a second larger triangle with the same theta labels the corresponding sides ky, kx, and kh" width=84 placement=none}
![Side ratios preserved by similarity](ratio-similarity.svg)
:::

## Investigation: same angle, different size

Before naming any trigonometric ratio, test what remains unchanged when a right triangle is enlarged.

Using a ruler and protractor, or dynamic-geometry software, construct three right triangles as follows. Draw a horizontal segment of the stated adjacent length. At its left endpoint, construct a ray making a $35^\circ$ angle above the segment. At the right endpoint, construct a perpendicular to the segment. The ray and perpendicular meet at the third vertex. Measure the opposite leg and hypotenuse to the nearest millimeter.

:::table{#tbl:ratio-investigation caption="Measured side lengths for three right triangles with a 35-degree reference angle"}
| Triangle | Adjacent (cm) | Opposite (cm) | Hypotenuse (cm) |
| --- | ---: | ---: | ---: |
| A | $4.0$ |  |  |
| B | $8.0$ |  |  |
| C | $12.0$ |  |  |
:::

::pagebreak[]

:::table{caption="Ratios calculated separately within each measured triangle"}
| Triangle | Opposite / hypotenuse | Adjacent / hypotenuse | Opposite / adjacent |
| --- | ---: | ---: | ---: |
| A |  |  |  |
| B |  |  |  |
| C |  |  |  |
:::

Compute each ratio from measurements in the same row and round ratios to three decimal places. Then answer:

1. When the adjacent leg doubles or triples, what happens approximately to the other two side lengths?
2. Which quantities vary from row to row? Which three quantities appear invariant?
3. The measurements contain construction and rounding error. Do close decimal results prove that the exact ratios are equal? Explain.
4. What geometric theorem could turn the numerical pattern into a proof for every such triangle?

> [!NOTE]
> **Conjecture.** Any two nondegenerate right triangles containing the same acute reference angle have equal opposite-to-hypotenuse, adjacent-to-hypotenuse, and opposite-to-adjacent ratios. Measurement supplies evidence for this statement; the argument below proves it.

## Why the ratios depend only on the angle

Take any two right triangles containing the same acute angle $\theta$. Each has a right angle and the angle $\theta$, so the triangles are similar by AA. Corresponding side lengths are proportional. For some scale factor $k>0$, the larger triangle has side lengths $kx$, $ky$, and $kh$. Therefore

$$
\frac{ky}{kh}=\frac{y}{h},\quad
\frac{kx}{kh}=\frac{x}{h},\quad
\frac{ky}{kx}=\frac{y}{x}.
$$

The scale factor cancels. Enlarging or shrinking the triangle changes its side lengths but not these ratios. Thus each ratio is a property of $\theta$, not of a particular drawing.

> [!IMPORTANT]
> Opposite and adjacent are relative names. If the other acute angle becomes the reference angle, the two legs exchange names. The hypotenuse does not.

::pagebreak[]

# The six trigonometric ratios

For an acute angle $\theta$ in a right triangle,

:::table{#tbl:six-ratios caption="Definitions of the six right-triangle trigonometric ratios"}
| Ratio | Definition using $x,y,h$ | Side-language form |
| --- | --- | --- |
| Sine | $\sin\theta=\dfrac{y}{h}$ | opposite / hypotenuse |
| Cosine | $\cos\theta=\dfrac{x}{h}$ | adjacent / hypotenuse |
| Tangent | $\tan\theta=\dfrac{y}{x}$ | opposite / adjacent |
| Cosecant | $\csc\theta=\dfrac{h}{y}$ | hypotenuse / opposite |
| Secant | $\sec\theta=\dfrac{h}{x}$ | hypotenuse / adjacent |
| Cotangent | $\cot\theta=\dfrac{x}{y}$ | adjacent / opposite |
:::

The familiar mnemonic **SOH-CAH-TOA** records only the first three definitions. It does not decide which angle is the reference angle, identify the hypotenuse, or justify any ratio.

Because $x,y,h>0$ for an acute angle, all six ratios are positive and defined. Also $0<y<h$ and $0<x<h$, so

$$0<\sin\theta<1,\quad 0<\cos\theta<1,$$

while tangent, cotangent, secant, and cosecant need not lie below $1$.

## Identities proved from the definitions

Dividing sine by cosine gives the quotient identity:

$$
\frac{\sin\theta}{\cos\theta}
=\frac{y/h}{x/h}
=\frac{y}{x}
=\tan\theta.
$$

Reversing a ratio proves each reciprocal identity:

$$
\csc\theta=\frac{1}{\sin\theta},\quad
\sec\theta=\frac{1}{\cos\theta},\quad
\cot\theta=\frac{1}{\tan\theta}.
$$

The side lengths satisfy $x^2+y^2=h^2$. Divide by $h^2$:

$$
\frac{x^2}{h^2}+\frac{y^2}{h^2}=1.
$$

Since $x/h=\cos\theta$ and $y/h=\sin\theta$,

:::equation{#eq:pythagorean-identity alt="sine squared theta plus cosine squared theta equals one" numbered=false}
\[
\sin^2\theta+\cos^2\theta=1.
\]
:::

This is the fundamental Pythagorean identity. Dividing it by $\cos^2\theta$ or $\sin^2\theta$ gives two further identities:

$$1+\tan^2\theta=\sec^2\theta,\quad 1+\cot^2\theta=\csc^2\theta.$$

# Exact values at $30^\circ$, $45^\circ$, and $60^\circ$

The exact values come from two elementary geometric constructions. They need not be accepted as an unexplained table.

## Proof of the $45^\circ$ values

Draw an isosceles right triangle with legs $1$ and $1$. Its two acute angles are equal, and together they measure $90^\circ$, so each is $45^\circ$. If the hypotenuse is $h$, the Pythagorean theorem gives

$$h^2=1^2+1^2=2,$$

and hence $h=\sqrt2$. Therefore the side ratio opposite : adjacent : hypotenuse is

$$1:1:\sqrt2.$$

It follows directly that

$$
\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2}=\frac{\sqrt2}{2},
\quad \tan45^\circ=1.
$$

Taking reciprocals gives

$$
\csc45^\circ=\sec45^\circ=\sqrt2,
\quad \cot45^\circ=1.
$$

## Proof of the $30^\circ$ and $60^\circ$ values

Start with an equilateral triangle of side $2$. Draw an altitude from its top vertex to the base. The two smaller triangles have equal hypotenuses, share the altitude, and have equal third sides inherited from the equilateral triangle; equivalently, their right-triangle hypotenuse-leg data make them congruent. Thus the altitude bisects both the base and the top $60^\circ$ angle.

Each half is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with hypotenuse $2$, short leg $1$, and altitude $a$. By Pythagoras,

$$a^2+1^2=2^2,$$

so $a^2=3$ and $a=\sqrt3$. The side ratio opposite $30^\circ$ : opposite $60^\circ$ : hypotenuse is therefore

$$1:\sqrt3:2.$$

:::figure{#fig:special caption="The two constructions that prove the exact side ratios for 45 degrees and for 30 and 60 degrees" alt="At left, an isosceles right triangle has legs 1 and 1, hypotenuse square root 2, and two 45-degree angles; at right, an equilateral triangle of side 2 is divided by an altitude into two triangles with sides 1, square root 3, and 2 and angles 30, 60, and 90 degrees" width=90 placement=none}
![Geometric proofs of the special-angle side ratios](special-angle-proofs.svg)
:::

Using $30^\circ$ as the reference angle,

$$
\sin30^\circ=\frac12,\quad
\cos30^\circ=\frac{\sqrt3}{2},\quad
\tan30^\circ=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}.
$$

Using $60^\circ$ as the reference angle exchanges the two legs:

$$
\sin60^\circ=\frac{\sqrt3}{2},\quad
\cos60^\circ=\frac12,\quad
\tan60^\circ=\sqrt3.
$$

The remaining values are reciprocals.

:::table{#tbl:exact-values caption="All six exact trigonometric values at the three special acute angles" summary="Rows give 30, 45, and 60 degrees; columns give sine, cosine, tangent, cosecant, secant, and cotangent in exact radical form."}
| $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ | $\csc\theta$ | $\sec\theta$ | $\cot\theta$ |
| --- | --- | --- | --- | --- | --- | --- |
| $30^\circ$ | $1/2$ | $\sqrt3/2$ | $\sqrt3/3$ | $2$ | $2\sqrt3/3$ | $\sqrt3$ |
| $45^\circ$ | $\sqrt2/2$ | $\sqrt2/2$ | $1$ | $\sqrt2$ | $\sqrt2$ | $1$ |
| $60^\circ$ | $\sqrt3/2$ | $1/2$ | $\sqrt3$ | $2\sqrt3/3$ | $2$ | $\sqrt3/3$ |
:::

> [!TIP]
> Reconstruct the table from the two triangles. A remembered table is fragile; the diagrams recover every entry and show which radicals belong to which angle.

## Complementary-angle identities

Let the other acute angle be $\phi=90^\circ-\theta$. The side opposite $\theta$ is adjacent to $\phi$, while the side adjacent to $\theta$ is opposite $\phi$. Hence

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

and similarly

$$
\tan\theta=\cot(90^\circ-\theta),\quad
\sec\theta=\csc(90^\circ-\theta).
$$

This proof also explains the symmetry of @tbl:exact-values. The $30^\circ$ and $60^\circ$ sine and cosine entries exchange places.

# Solving right triangles

To **solve a right triangle** is to determine every unknown side and angle. Mark the reference angle first. Then choose a ratio containing the known side and the required side. Write the ratio equation before entering numbers.

## Finding a side

**Example 1: tangent for a leg.** A right triangle has an acute angle of $38^\circ$ and adjacent leg $12.0$ cm. If the opposite leg is $y$, then

$$\tan38^\circ=\frac{y}{12.0}.$$

Therefore

$$y=12.0\tan38^\circ\approx9.38\text{ cm}.$$

The other acute angle is $52^\circ$. Pythagoras gives a hypotenuse of approximately $15.23$ cm, which is longer than both legs.

**Example 2: cosine for the hypotenuse.** A guy wire makes a $64^\circ$ angle with level ground. Its anchor is $7.5$ m from the pole. If the wire length is $h$, then

$$\cos64^\circ=\frac{7.5}{h},$$

so

$$h=\frac{7.5}{\cos64^\circ}\approx17.11\text{ m}.$$

## Finding an angle

If a ratio is known, an inverse trigonometric function returns the acute angle. For example,

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

means "the acute angle whose sine is $0.6$." It does **not** mean $1/\sin(0.6)$; the reciprocal of sine is written $\csc$.

**Example 3.** A right triangle has opposite leg $9$ and adjacent leg $14$. Then

$$\tan\theta=\frac{9}{14},$$

and in degree mode,

$$\theta=\tan^{-1}\left(\frac{9}{14}\right)\approx32.74^\circ.$$

The second acute angle is $90^\circ-32.74^\circ=57.26^\circ$.

Why is the acute answer unique? Fix the adjacent leg at $1$. As an acute ray rotates upward from $0^\circ$ toward $90^\circ$, its height where it meets the line $x=1$ increases continuously from $0$ without bound. That height is $\tan\theta$. Thus each positive tangent ratio corresponds to exactly one acute angle. The inverse tangent names that angle.

## A complete triangle

**Example 4.** A right triangle has hypotenuse $20$ ft and one acute angle $27^\circ$. Let $x$ be adjacent and $y$ opposite to $27^\circ$.

$$
x=20\cos27^\circ\approx17.82\text{ ft},
\quad
y=20\sin27^\circ\approx9.08\text{ ft}.
$$

The remaining angle is $63^\circ$. A check using unrounded values gives

$$x^2+y^2=20^2(\cos^227^\circ+\sin^227^\circ)=400.$$

> [!WARNING]
> Set the calculator to degree mode. Keep unrounded values during intermediate work and round once at the end. A diagram may be schematic and need not be drawn to scale.

# Heights and distances

An **angle of elevation** is measured upward from a horizontal line. An **angle of depression** is measured downward from a horizontal line. When the observer's horizontal and the ground are parallel, an angle of depression equals the corresponding angle of elevation by alternate interior angles.

:::figure{#fig:elevation caption="Angles of elevation and depression are measured from parallel horizontal lines" alt="A vertical tower and level ground form a right angle; a sloping line of sight connects an observer on the ground to the tower top, and a second observer above views a ground point downward; marked alternate interior angles show why elevation and depression use parallel horizontals" width=88 placement=none}
![Angles of elevation and depression](elevation-depression.svg)
:::

## One observation of a height

**Example 5: building with observer height.** From level ground, an observer whose eye is $1.65$ m above the ground stands $28.0$ m from a building. The angle of elevation to the roof is $41^\circ$. The vertical rise from eye level to the roof is

$$r=28.0\tan41^\circ\approx24.34\text{ m}.$$

The building height is therefore

$$H=r+1.65\approx25.99\text{ m},$$

or $26.0$ m to the nearest tenth. The observer height is added after the right-triangle rise has been found.

## Angle of depression

**Example 6: lighthouse.** From the top of a $36$ m lighthouse, the angle of depression to a boat is $18^\circ$. Level sea and the horizontal through the observer are parallel, so the boat's angle of elevation is also $18^\circ$. If $d$ is the horizontal distance,

$$\tan18^\circ=\frac{36}{d},$$

and hence

$$d=\frac{36}{\tan18^\circ}\approx110.80\text{ m}.$$

## Two observations of an inaccessible height

**Example 7.** Two points $A$ and $B$ lie on level ground on the same side of a vertical tower. Point $B$ is $30$ m closer to the tower. The angles of elevation are $35^\circ$ at $A$ and $52^\circ$ at $B$. Let $x$ be the distance from $B$ to the tower and $H$ the tower height. Then

$$H=x\tan52^\circ=(x+30)\tan35^\circ.$$

Therefore

$$
x=\frac{30\tan35^\circ}{\tan52^\circ-\tan35^\circ}
\approx36.24\text{ m},
$$

and

$$H=x\tan52^\circ\approx46.38\text{ m}.$$

The two equations describe the same height from different observation points. This equality is the essential model.

## Modeling assumptions

Real measurements carry uncertainty. State whether the object is vertical, the ground is level, and the recorded angle is measured from a horizontal. For a distant target, distinguish horizontal distance from sloping line-of-sight distance. A calculated answer should not claim more precision than the measurements support.

# Practice

Give exact values where possible. For measured data, keep guard digits and round the final answer as directed.

::::regions{tracks="1 1" gutter=18}
:::region
**A. Meaning and proof**

1. Relative to angle $\theta$ in @fig:ratios, identify the opposite side, adjacent side, and hypotenuse.
2. Explain why two right triangles containing the same acute angle are similar.
3. If every side of a right triangle is tripled, prove that its sine, cosine, and tangent ratios are unchanged.
4. A student labels a leg "the opposite" without marking a reference angle. Explain what is missing.
5. From the definitions, prove $\tan\theta=\sin\theta/\cos\theta$.
6. From $x^2+y^2=h^2$, prove $\sin^2\theta+\cos^2\theta=1$.

**B. Exact values and identities**

7. Reconstruct the $45^\circ$ triangle and derive $\sin45^\circ$ exactly.
8. In the equilateral-triangle proof, explain why the altitude creates a segment of length $1$ when each original side is $2$.
9. Derive $\cos30^\circ$ and $\tan30^\circ$ without using a calculator.
10. Derive $\sin60^\circ$ and $\sec60^\circ$ without using a calculator.
11. Evaluate exactly: $4\sin30^\circ-2\cos60^\circ$.
12. Evaluate exactly: $\tan30^\circ\tan60^\circ$.
13. Evaluate exactly: $\sin^245^\circ+\cos^245^\circ$.
14. If $\sin\theta=5/13$ for acute $\theta$, find the other five ratios exactly.
15. Prove $1+\tan^2\theta=\sec^2\theta$ from the fundamental Pythagorean identity.
:::
:::region
**C. Solve right triangles**

16. A right triangle has $\theta=36^\circ$ and adjacent leg $10$ cm. Find the opposite leg to the nearest tenth.
17. A right triangle has $\theta=54^\circ$ and opposite leg $12$ m. Find the hypotenuse to the nearest tenth.
18. A right triangle has legs $8$ and $15$. Find both acute angles to the nearest tenth of a degree.
19. A right triangle has hypotenuse $26$ and adjacent leg $10$. Find $\theta$ and the opposite leg.
20. Solve a right triangle with hypotenuse $18$ and one acute angle $41^\circ$.
21. Explain why $\sin^{-1}x$ and $1/\sin x$ are not the same operation.

**D. Applications and diagnosis**

22. From $24$ m away, the angle of elevation to the top of a vertical flagpole is $37^\circ$. Find its height to the nearest tenth, ignoring observer height.
23. A $6.5$ m ladder makes a $72^\circ$ angle with level ground. How high up a wall does it reach?
24. A wheelchair ramp rises $0.75$ m over a horizontal run of $9.0$ m. Find its angle of inclination to the nearest tenth of a degree.
25. From the top of a $48$ m cliff, the angle of depression to a boat is $23^\circ$. Find the horizontal distance to the boat.
26. A surveyor's eye is $1.60$ m above level ground. From $42$ m away, the angle of elevation to a tower top is $29^\circ$. Find the tower height.
27. Two observation points on level ground are $20$ m apart and lie on the same side of a tower. Their angles of elevation are $28^\circ$ and $43^\circ$, with the larger angle at the nearer point. Find the tower height.
28. A student computes $12/\tan35^\circ$ when the diagram requires the side opposite $35^\circ$ and gives adjacent side $12$. Diagnose and correct the equation.
29. A calculator reports $\sin30= -0.988$. Identify the likely setting error and give the exact value.
30. **Exit ticket.** Prove $\sin\theta=\cos(90^\circ-\theta)$, then use it to evaluate $\cos60^\circ$ from $\sin30^\circ$.
:::
::::

## Working record

For each numerical model, include a labeled sketch, the selected ratio before substitution, unrounded calculator input, final units, and a reasonableness check.

:::keep-lines{count=8}
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:::

## Sources

Curriculum and mathematical references are listed in `references.bib`. The exposition, exercises, and diagrams in this lesson are newly written; classical sources were consulted for mathematical scope, not copied.