# Right triangle trigonometry

> IB Mathematics Analysis and Approaches HL · Geometry & Trigonometry
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u3-right-triangle-trigonometry/

This module covers core trigonometric ratios for right-angled triangles, the foundation for all further trigonometry in IB AA HL. You will learn to solve for unknown sides and angles, and apply these skills to common 2D and 3D exam problems.

**Prerequisites:** [Pythagoras' theorem](https://www.owlsprep.com/study/ib-math-aa-hl-algebra-pythagoras-theorem/); Properties of right triangles and parallel lines

## Learning objectives

- Define sine, cosine and tangent ratios for acute angles in right triangles
- Solve for unknown sides and angles in right triangles
- Apply right triangle trigonometry to 2D and 3D geometry problems
- Select the correct trigonometric ratio for exam problems

## Defining Trigonometric Ratios

**Trigonometric Ratios** — Ratios of side lengths for an acute angle $\theta$ in a right-angled triangle, that relate the size of the angle to the lengths of the triangle's sides

*Notation:* $\sin\theta$, $\cos\theta$, $\tan\theta$

*Example:* Ratios are always defined relative to the acute angle of interest, not the right angle

> **SOH-CAH-TOA Memory Hook**
>
> SOH: Sine = Opposite / Hypotenuse  
> CAH: Cosine = Adjacent / Hypotenuse  
> TOA: Tangent = Opposite / Adjacent

**Worked example:** A right triangle has hypotenuse 10 cm and an acute angle of $30^\circ$. Find the length of the side opposite $30^\circ$.

1. We know the hypotenuse and need the opposite side, so we use the sine ratio:
2. $$\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{10}$$
3. Substitute the known value of $\sin 30^\circ = 0.5$ and rearrange:
4. $$0.5 = \frac{x}{10} \implies x = 10 \times 0.5 = 5$$
5. The opposite side length is 5 cm.

## Solving for Unknown Sides

When you know one acute angle and one side length, you can use the appropriate trig ratio to find any other side. The most important step is correctly labeling the sides **relative to the known angle** to select the right ratio.

**Worked example:** A right triangle has an acute angle of $45^\circ$, and the side adjacent to this angle is 7 cm. Find the length of the hypotenuse.

1. Label the sides: the known side is adjacent to $45^\circ$, and we need the hypotenuse, so we use cosine:
2. $$\cos 45^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{h}$$
3. Rearrange to isolate the unknown hypotenuse $h$:
4. $$h = \frac{7}{\cos 45^\circ}$$
5. Substitute $\cos 45^\circ = \frac{\sqrt{2}}{2}$ and simplify:
6. $$h = \frac{7}{\frac{\sqrt{2}}{2}} = 7\sqrt{2} \approx 9.90 \text{ cm}$$

**Check your understanding**

Check your ratio selection skill:

1. You know the length of the opposite side and hypotenuse, and need to find the angle. Which ratio do you use?

   - Sine
   - Cosine
   - Tangent
   - Pythagoras' theorem

   *Why:* Correct: Sine is defined as opposite over hypotenuse, which matches the given sides.

## Solving for Unknown Angles

When you know two side lengths, you can use inverse trigonometric functions (written $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$) to find the measure of an unknown acute angle. Inverse functions take a ratio value and return the corresponding angle.

**Worked example:** A 3-4-5 right triangle has its right angle between the sides of 3 cm and 4 cm. Find the angle opposite the 3 cm side.

1. The side opposite the unknown angle is 3 cm, the adjacent side is 4 cm, so we use tangent:
2. $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4} = 0.75$$
3. Apply the inverse tangent function to both sides to solve for $\theta$:
4. $$\theta = \tan^{-1}(0.75) \approx 36.9^\circ$$
5. Verify: The other acute angle is $90^\circ - 36.9^\circ = 53.1^\circ$, which matches $\tan^{-1}(\frac{4}{3}) \approx 53.1^\circ$, so the answer is correct.

## Applications to 3D Geometry

A common HL exam question asks for angles or lengths in 3D shapes like cuboids, pyramids and prisms. The core strategy is to identify a right triangle within the 3D shape that contains your unknown value, then apply right triangle trigonometry to that 2D triangle.

> **3D Problem Strategy**
>
> Always draw a separate 2D diagram of the right triangle you are using, to avoid mislabeling sides in the 3D drawing.

**Worked example:** A cuboid has length 5 cm, width 4 cm, height 3 cm. Find the angle between the space diagonal of the cuboid and the base of the cuboid.

1. First calculate the diagonal of the base rectangle, which is the adjacent side of our right triangle:
2. $$\text{base diagonal} = \sqrt{5^2 + 4^2} = \sqrt{41} \approx 6.403 \text{ cm}$$
3. The right triangle has opposite side equal to the height of the cuboid (3 cm) and adjacent side equal to the base diagonal. We use tangent to find the angle $\theta$:
4. $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{\sqrt{41}} \approx 0.4685$$
5. Apply inverse tangent to get the angle:
6. $$\theta = \tan^{-1}(0.4685) \approx 25.1^\circ$$

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Labeling sides relative to the right angle instead of the acute angle of interest
  - Why it fails: All trigonometric ratios are defined relative to the acute angle you are working with
  - Correct: Always label opposite and adjacent relative to the acute angle you know or are trying to find
- **Wrong:** Working in radians mode on your calculator for degree-based angle problems
  - Why it fails: Most calculators default to radians for calculus problems, leading to incorrect numerical values
  - Correct: Always check your calculator is in degrees mode before starting right triangle trig problems
- **Wrong:** Skipping drawing a separate 2D triangle for 3D problems
  - Why it fails: 3D perspective drawings make it easy to misidentify right angles and side labels
  - Correct: Always draw a clear 2D diagram of the right triangle you are using for 3D problems
- **Wrong:** Mixing up the numerator and denominator when writing the ratio
  - Why it fails: Rushing to write the equation without double checking the SOH-CAH-TOA rule
  - Correct: After writing the ratio, confirm it matches the SOH-CAH-TOA mnemonic before solving
- **Wrong:** Using trigonometry when Pythagoras' theorem is sufficient to find an unknown side
  - Why it fails: Confusing when to use Pythagoras vs trig when you already know two sides and need the third
  - Correct: Use Pythagoras' theorem when you know two sides of a right triangle and need the third, use trig when you know an angle

## Cheatsheet

| Ratio | Formula | Use for unknown angles |
| --- | --- | --- |
| Sine $\sin\theta$ | $\frac{\text{Opposite}}{\text{Hypotenuse}}$ | $\theta = \sin^{-1}\left(\frac{\text{Opp}}{\text{Hyp}}\right)$ |
| Cosine $\cos\theta$ | $\frac{\text{Adjacent}}{\text{Hypotenuse}}$ | $\theta = \cos^{-1}\left(\frac{\text{Adj}}{\text{Hyp}}\right)$ |
| Tangent $\tan\theta$ | $\frac{\text{Opposite}}{\text{Adjacent}}$ | $\theta = \tan^{-1}\left(\frac{\text{Opp}}{\text{Adj}}\right)$ |
| Mnemonic | SOH-CAH-TOA | Label relative to $\theta$ |

## What's next

Right triangle trigonometry is the foundation for all further trigonometry in IB AA HL. Next, you will extend your understanding of trigonometric ratios to all angles (not just acute angles in right triangles) using the unit circle, which allows you to model periodic functions and solve trigonometric equations. Right triangle trigonometry is also heavily used in non-right triangle trigonometry (the sine and cosine rules) and forms the basis of most 3D geometry problems that appear regularly in Paper 1 and Paper 2 exams. Mastery of this basic sub-topic is essential to avoid losing easy marks on multi-step exam questions.

- [Non-right triangle trigonometry](https://www.owlsprep.com/study/ib-math-aa-hl-u3-non-right-triangle-trigonometry/)
- [Trigonometric equations](https://www.owlsprep.com/study/ib-math-aa-hl-u3-trigonometric-equations/)
- [Compound and double angle identities](https://www.owlsprep.com/study/ib-math-aa-hl-u3-compound-and-double-angle-identities/)

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