# Right-Angled Trigonometry

> Pythagoras, sine, cosine and tangent, angles of elevation, bearings and exact values.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/right-angled-trigonometry/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The tree and the shadow
Jack wants to know how tall the gum tree in his backyard is, but he cannot climb it. He walks 20 m away from the trunk and uses a phone app to measure the angle up to the top: $35^\circ$.

With one length and one angle, trigonometry gives the height without a ladder. Surveyors, pilots and builders do this every day. The idea goes back thousands of years to people measuring pyramids by their shadows.

**In short:** Right-angled trigonometry links the angles and sides of a right-angled triangle with SOH CAH TOA: $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$. Inverse trig finds an angle from two sides.

## Starting point: Years 9 and 10

### Pythagoras' theorem

In a right-angled triangle with hypotenuse $c$ (the longest side, opposite the right angle):

$$a^2 + b^2 = c^2$$

So a triangle with sides 6 and 8 has hypotenuse $\sqrt{36 + 64} = 10$.

### Naming the sides

Pick an angle $\theta$. Then:
- **Hypotenuse (H):** the longest side, opposite the right angle.
- **Opposite (O):** across from $\theta$.
- **Adjacent (A):** next to $\theta$, not the hypotenuse.

*Diagram: Right-angled triangle with angle theta, showing the hypotenuse, opposite and adjacent sides*
*The opposite and adjacent sides depend on which angle you pick.*

### The three ratios

### Key idea: SOH CAH TOA
$$\sin \theta = \frac{O}{H} \qquad \cos \theta = \frac{A}{H} \qquad \tan \theta = \frac{O}{A}$$

### Worked example: Back to the gum tree
Jack is 20 m from the tree (adjacent) and wants the height (opposite). Use tan.

$$\tan 35^\circ = \frac{h}{20} \;\Rightarrow\; h = 20 \tan 35^\circ \approx 14.0 \text{ m}$$

### Worked example: Finding an angle
A ramp rises 1.2 m over a horizontal distance of 8 m. What angle does it make with the ground?

$$\tan \theta = \frac{1.2}{8} = 0.15 \;\Rightarrow\; \theta = \tan^{-1}(0.15) \approx 8.5^\circ$$

### Calculator tip: Degree mode, and finding angles
Check that the screen shows `D` (degrees) before any trig question. In radian mode, $\sin 30$ gives $-0.988$ instead of 0.5.

To find an angle, use the inverse key: `SHIFT sin` is $\sin^{-1}$. To change a decimal angle into degrees and minutes, press the `° ' "` key after the answer: 36.87° becomes 36°52'.

**One ratio gives the others.** If $\tan\theta = \dfrac{3}{4}$ and $\theta$ is acute, draw a right triangle with opposite 3 and adjacent 4. Pythagoras gives the hypotenuse 5, so $\sin\theta = \dfrac{3}{5}$ and $\cos\theta = \dfrac{4}{5}$.

### Angles of elevation and depression

The **angle of elevation** is measured **up** from the horizontal. The **angle of depression** is measured **down** from the horizontal. They are equal for two people looking at each other, because they are alternate angles between parallel lines.

*Diagram: A lookout on top of a cliff sees a boat on the sea. The angle of depression is measured down from the horizontal at the lookout, and the angle of elevation is measured up from the sea at the boat. Both angles are theta*
*Both angles are measured from a horizontal line, never from the cliff. The horizontal at the top and the sea are parallel, so the two angles are equal alternate angles.*

**Eye height.** An angle of elevation measured from someone's eyes gives the height above their eyes. Add their eye height at the end: if the triangle gives 26.2 m and the eyes are 1.6 m above the ground, the tree is 27.8 m tall.

*Diagram: A person looks up at the top of a tree. The angle of elevation is at their eyes, 1.6 m above the ground. The right-angled triangle gives 26.2 m from eye level to the top, so the tree is 26.2 plus 1.6, which is 27.8 m tall*
*Trigonometry gives the height above the eyes (26.2 m). The eyes are 1.6 m above the ground, so add that at the end: 27.8 m.*

### Bearings

A **true bearing** is a three-digit angle measured **clockwise from north**. East is $090^\circ$, south is $180^\circ$, west is $270^\circ$. A **compass bearing** like S$40^\circ$W means start facing south and turn $40^\circ$ toward west.

To change a compass bearing to a true bearing, start from the true bearing of the first direction. S$40^\circ$W is $180^\circ + 40^\circ = 220^\circ$. S$50^\circ$E is $180^\circ - 50^\circ = 130^\circ$. N$30^\circ$W is $360^\circ - 30^\circ = 330^\circ$.

### Watch out: Calculator mode
Check your calculator is in **degrees** for these problems. A wrong mode is the most common lost mark in trigonometry.

### Exam tip: Draw it first
Draw a diagram, mark the right angle, label O, A and H from the angle you are using, then choose the ratio.

**Mind map**

```text
Right-angled trig
  Pythagoras
    a² + b² = c²
  Sides
    Hypotenuse
    Opposite
    Adjacent
  Ratios
    sin = O/H
    cos = A/H
    tan = O/A
  Uses
    Elevation and depression
    Bearings
  Exact values
    45° triangle
    30° and 60° triangle
```

### Year 11: Exact values
Two special triangles give exact ratios with no calculator.

**Half a square:** sides 1, 1, $\sqrt{2}$ with angles $45^\circ$, $45^\circ$, $90^\circ$.

**Half an equilateral triangle:** sides 1, $\sqrt{3}$, 2 with angles $30^\circ$, $60^\circ$, $90^\circ$.

*Diagram: Two special triangles. Left: half of a unit square, with legs 1 and 1, hypotenuse root 2 and two 45 degree angles. Right: half of an equilateral triangle of side 2, with sides 1, root 3 and 2, and angles 30, 60 and 90 degrees*
*Cut a square of side 1 along its diagonal, or an equilateral triangle of side 2 down its middle. Pythagoras gives the missing sides: √(1² + 1²) = √2 and √(2² − 1²) = √3.*

| | $30^\circ$ | $45^\circ$ | $60^\circ$ |
|---|---|---|---|
| sin | $\dfrac{1}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{\sqrt{3}}{2}$ |
| cos | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{1}{2}$ |
| tan | $\dfrac{1}{\sqrt{3}}$ | 1 | $\sqrt{3}$ |

Notice that $\sin 30^\circ = \cos 60^\circ$. In general $\sin \theta = \cos(90^\circ - \theta)$. These are **complementary** angles.

### Year 11: Problems in three dimensions
In 3D questions, find a right-angled triangle hiding in the shape and solve it, then use that answer in a second triangle.

**Example.** A box is 4 by 3 by 12. The floor diagonal is $\sqrt{4^2 + 3^2} = 5$. The space diagonal is $\sqrt{5^2 + 12^2} = 13$. The angle it makes with the floor is $\tan^{-1}\left(\dfrac{12}{5}\right) \approx 67.4^\circ$.

*Diagram: A box 4 long, 3 wide and 12 high. The floor diagonal of length 5 and the upright edge of 12 make a right-angled triangle with the space diagonal of length 13, which makes angle theta with the floor*
*First the floor triangle: 4, 3 and the floor diagonal 5. Then the upright triangle: 5 along the floor, 12 up the edge, and the space diagonal 13. The angle θ is in the upright triangle.*

**Two sightings.** From $A$ the top of a tower has elevation $30^\circ$; walking 40 m closer to $B$, it is $60^\circ$. With height $h$: from $B$, the distance to the tower is $\dfrac{h}{\tan 60^\circ} = \dfrac{h}{\sqrt{3}}$, and from $A$ it is $\dfrac{h}{\tan 30^\circ} = \sqrt{3}h$. The difference is 40: $\sqrt{3}h - \dfrac{h}{\sqrt{3}} = \dfrac{2h}{\sqrt{3}} = 40$, so $h = 20\sqrt{3}$ m exactly.

*Diagram: A tower of height h. From A the angle of elevation of the top is 30 degrees, and from B, 40 m closer, it is 60 degrees*
*Two right-angled triangles share the height h. The distance from A is √3h and from B it is h ÷ √3. Their difference is the 40 m walked.*

**Bearings in 3D.** If $C$ is due south of the foot of a tower and $D$ is due west, then angle $CBD$ at the foot is $90^\circ$. Find $BC$ and $BD$ from the elevations, then $CD$ by Pythagoras on the ground. A point 30 m south and 40 m east of the foot is $\sqrt{30^2 + 40^2} = 50$ m away.

*Diagram: A tower with foot B. Point D is due west of the foot and point C is due south, so angle CBD on the ground is 90 degrees. The elevation of the top is alpha from D and beta from C, and CD is found by Pythagoras on the ground*
*Each line of sight makes an upright right-angled triangle with the tower: use α to find BD and β to find BC. West and south are at right angles, so the shaded ground triangle gives CD by Pythagoras.*

### Year 12: Exact values in calculus
Year 12 calculus uses radians, but the exact values are the same: $\dfrac{\pi}{6} = 30^\circ$, $\dfrac{\pi}{4} = 45^\circ$, $\dfrac{\pi}{3} = 60^\circ$. For example,

$$\int_0^{\frac{\pi}{6}} \cos x \, dx = \sin \frac{\pi}{6} - \sin 0 = \frac{1}{2}$$

Learn the table now and it saves time in every calculus exam.

## Quick summary

- Pythagoras: $a^2 + b^2 = c^2$, with $c$ the hypotenuse.
- SOH CAH TOA links an angle to two sides.
- Use inverse trig ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$) to find angles.
- Elevation is up from the horizontal, depression is down. True bearings go clockwise from north.
- Exact values come from the $45^\circ$ and $30^\circ$-$60^\circ$ triangles.

## Quiz

Try these questions from the chapter quiz.

1. A right-angled triangle has shorter sides 5 and 12. The hypotenuse is:
   - A. 17
   - B. 13
   - C. 7
   - D. 169
   - Answer: B. 13. $\sqrt{25 + 144} = \sqrt{169} = 13$.

2. Which ratio is $\dfrac{\text{opposite}}{\text{hypotenuse}}$?
   - A. $\sin \theta$
   - B. $\cos \theta$
   - C. $\tan \theta$
   - D. $\theta$
   - Answer: A. $\sin \theta$. SOH.

3. A ladder 5 m long leans at $60^\circ$ to the ground. How high up the wall does it reach?
   - A. 2.5 m
   - B. 4.33 m
   - C. 5.77 m
   - D. 8.66 m
   - Answer: B. 4.33 m. $h = 5 \sin 60^\circ \approx 4.33$ m.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/right-angled-trigonometry/#quiz