# Right-angled triangle trigonometry

> IB Mathematics: Applications and Interpretation SL · Unit 3: Geometry and Trigonometry
> Source: https://www.owlsprep.com/study/ib-math-ai-sl-u3-right-angled-triangle-trigonometry/

This sub-topic introduces core trigonometric ratios for right-angled triangles, the foundation for all further trigonometry in IB AI SL. You will learn to find unknown sides and angles, and apply these skills to real-world problems including bearings and elevation.

**Prerequisites:** [Pythagoras' theorem for right-angled triangles](https://www.owlsprep.com/study/ib-math-ai-sl-u3-pythagoras-theorem/)

## Learning objectives

- Identify sides of a right-angled triangle relative to an acute angle
- Apply SOHCAHTOA to find missing sides and angles
- Solve practical problems involving right-angled triangles
- Apply trigonometry to problems involving elevation, depression and bearings

## 1. Identifying sides of a right-angled triangle

**Sides of a right-angled triangle** — All right-angled triangles have three sides, defined relative to any given acute angle $\theta$: the hypotenuse is always the longest side, opposite the right angle. The opposite side is across from $\theta$, and the adjacent side is next to $\theta$ (not the hypotenuse).

*Example:* For triangle $ABC$ right-angled at $C$, with $\theta$ at $A$: hypotenuse = $AB$, opposite = $BC$, adjacent = $AC$

**Worked example:** In right-angled triangle $XYZ$, right-angled at $Y$, with acute angle $30^\circ$ at $X$. Identify hypotenuse, opposite and adjacent sides relative to the $30^\circ$ angle.

1. Step 1: Locate the right angle at $Y$. The hypotenuse is always opposite the right angle, so hypotenuse = $XZ$.
2. Step 2: The given angle is at $X$, so the side across from $X$ is $YZ$. This is the opposite side.
3. Step 3: The remaining side next to $X$ (not the hypotenuse) is $XY$. This is the adjacent side.

## 2. Trigonometric ratios and SOHCAHTOA

**Trigonometric ratios** — The three core trigonometric ratios are defined as ratios of side lengths of a right-angled triangle:

*Notation:* For acute angle $\theta$:

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

> **SOH-CAH-TOA**
>
> **S**ine = **O**pposite / **H**ypotenuse (SOH)  
> **C**osine = **A**djacent / **H**ypotenuse (CAH)  
> **T**angent = **O**pposite / **A**djacent (TOA)

**Worked example:** A right-angled triangle has an acute angle of $25^\circ$ and hypotenuse 12 cm. Find the length of the side opposite the $25^\circ$ angle, correct to 2 decimal places.

1. Step 1: We know the hypotenuse and need the opposite side. This matches the sine ratio from SOHCAHTOA.
2. Step 2: Write the sine ratio, let the unknown opposite side be $x$:
3. $$\sin(25^\circ) = \frac{x}{12}$$
4. Step 3: Rearrange to isolate $x$:
5. $$x = 12 \times \sin(25^\circ)$$
6. Step 4: Calculate with a calculator in degree mode: $x \approx 12 \times 0.4226 = 5.07$ cm

> **Exam tip:** Always confirm your calculator is in degree mode before starting any trigonometry calculation for IB exams.

*Calculator:* allowed

## 3. Calculating unknown acute angles

When you know two side lengths of a right-angled triangle, you can calculate an unknown acute angle using inverse trigonometric functions. These functions reverse the action of sine, cosine and tangent, returning the angle value from a given ratio.

**Inverse trigonometric functions** — If $\sin\theta = k$, then $\theta = \sin^{-1}(k)$ gives the angle whose sine ratio equals $k$. This works identically for inverse cosine and tangent.

*Notation:* $\theta = \sin^{-1}(k), \theta = \cos^{-1}(k), \theta = \tan^{-1}(k)$

**Worked example:** Find the unknown acute angle $\theta$ in a right-angled triangle, where the side adjacent to $\theta$ is 7 cm and the hypotenuse is 10 cm. Give your answer to the nearest degree.

1. Step 1: We know adjacent and hypotenuse, so we use the cosine ratio.
2. Step 2: Write the ratio:
3. $$\cos\theta = \frac{7}{10} = 0.7$$
4. Step 3: Apply inverse cosine to both sides to isolate $\theta$:
5. $$\theta = \cos^{-1}(0.7)$$
6. Step 4: Calculate with a calculator: $\theta \approx 46^\circ$

*Calculator:* allowed

## 4. Applied right-angled trigonometry

Right-angled trigonometry is commonly used to solve real-world problems including finding heights of objects, distances between points, angles of elevation/depression, and bearings. Always start by drawing a labelled diagram to model the problem.

**Worked example:** The angle of elevation from a point 50 m horizontally from the base of a tree to the top of the tree is $22^\circ$. Find the height of the tree, correct to 1 decimal place.

1. Step 1: Draw a right-angled triangle: the horizontal distance (50 m) is adjacent to the $22^\circ$ angle, and the height $h$ of the tree is opposite the angle.
2. Step 2: We have adjacent, need opposite, so use tangent:
3. $$\tan(22^\circ) = \frac{h}{50}$$
4. Step 3: Rearrange for $h$:
5. $$h = 50 \times \tan(22^\circ)$$
6. Step 4: Calculate: $h \approx 50 \times 0.4040 = 20.2$ m

> **Exam tip:** Always label your diagram clearly with all given information to avoid mixing up opposite and adjacent sides.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Mixing up opposite and adjacent sides relative to the given angle
  - Why it fails: Opposite and adjacent are defined relative to the acute angle you are working with, not the right angle
  - Correct: Always re-label the triangle from the perspective of the angle you are using
- **Wrong:** Leaving your calculator in radian mode instead of degree mode
  - Why it fails: IB AI SL almost exclusively uses degrees for trigonometry questions, leading to incorrect answers
  - Correct: Check your calculator mode before starting any trigonometry calculation
- **Wrong:** Using the regular trigonometric function instead of inverse when finding an angle
  - Why it fails: A regular trig function takes an angle and gives a ratio; you need the inverse to get an angle from a ratio
  - Correct: Always use $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$ when calculating an unknown angle
- **Wrong:** Using Pythagoras' theorem when an angle is given
  - Why it fails: Pythagoras only works when you know two sides and have no angle information
  - Correct: If you have an angle, always use trigonometric ratios to find the unknown side

## Cheatsheet

| Concept | Rule | Notation |
| --- | --- | --- |
| Hypotenuse | Longest side, opposite right angle | $c$ |
| Opposite side | Side across from angle $\theta$ | $O$ |
| Adjacent side | Side next to angle $\theta$ | $A$ |
| Sine | SOH = $O / H$ | $\sin\theta = O/H$ |
| Cosine | CAH = $A / H$ | $\cos\theta = A/H$ |
| Tangent | TOA = $O / A$ | $\tan\theta = O/A$ |
| Find angle | Use inverse trig function | $\theta = \sin^{-1}(O/H)$ |
| Find side | Rearrange ratio to isolate unknown | $O = H \times \sin\theta$ |

## What's next

Right-angled triangle trigonometry is the foundation for all further trigonometry in IB AI SL. Next, you will extend these ideas to solve non-right angled triangles using the sine and cosine rules, which are required for most irregular triangle problems in exams. You will also use trigonometry to model periodic phenomena like tide heights and seasonal temperatures later in the course, so mastering SOHCAHTOA is critical for success in these more advanced topics.

- [Sine rule, cosine rule for non-right triangles](https://www.owlsprep.com/study/ib-math-ai-sl-u3-sine-rule-cosine-rule-for/)
- [Bearings and trigonometric applications](https://www.owlsprep.com/study/ib-math-ai-sl-u3-bearings-and-trigonometric-applications/)
- [Sinusoidal functions: amplitude, period, translation](https://www.owlsprep.com/study/ib-math-ai-sl-u3-sinusoidal-functions-amplitude-period-translation/)

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