# Trigonometry in Three Dimensions

> Solve problems in 3D by finding the right triangles inside boxes, pyramids and hills, with angles of elevation and bearings.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/three-dimensional-trigonometry/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The drone pilot
Sam flies a drone above a football field. From the corner flag, the drone is 40 m east, 30 m north and 25 m up. How far is the drone from Sam in a straight line, and at what angle must Sam look up to see it?

The answer lives in three dimensions, but the trick is simple: find flat right-angled triangles hidden inside the 3D picture and solve them one at a time.

**In short:** Trigonometry in three dimensions means finding right-angled triangles inside a 3D shape and solving them one at a time. A vertical line is at right angles to every line on the ground, and the diagonal of a box with sides $a$, $b$ and $c$ is $\sqrt{a^2 + b^2 + c^2}$.

## Starting point: what you already know

- **Pythagoras:** $c^2 = a^2 + b^2$ in a right-angled triangle.
- **SOH CAH TOA** to find sides and angles.
- **Sine rule** $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ and **cosine rule** $c^2 = a^2 + b^2 - 2ab\cos C$ for triangles without a right angle.
- **Angle of elevation:** the angle you look **up** from the horizontal. **Angle of depression:** the angle you look **down**.
- **True bearings** are measured clockwise from north, written with three digits, like $060^\circ$.

## The method

1. Draw a clear 3D sketch. Label every point with a capital letter.
2. Mark right angles. A vertical line meets any horizontal line at $90^\circ$.
3. Find a triangle that contains what you want and enough known lengths or angles.
4. If it needs a length you do not have, find that length first from another triangle.
5. Redraw each triangle flat, then solve it.

### Key idea: The golden rule
Vertical edges are at right angles to **everything** on the ground. That is where the hidden right angles come from.

### Calculator tip: Store each length
In three-dimensional problems one triangle feeds the next. Store each length with `STO` (in `A`, `B`, `C` and so on) instead of writing down a rounded value, so the final angle stays accurate. Check you are in degree mode.

## The rectangular box

%%FIG:box3d%%

### Worked example: The space diagonal
The box is 8 long, 6 deep and 5 high. Find $AG$ and the angle $\theta$ it makes with the floor.

**Floor triangle** $ABC$, with the right angle at $B$:
$$AC^2 = 8^2 + 6^2 = 100 \quad\Rightarrow\quad AC = 10$$

**Upright triangle** $ACG$, with the right angle at $C$ because $CG$ is vertical:
$$AG^2 = 10^2 + 5^2 = 125 \quad\Rightarrow\quad AG = 5\sqrt{5} \approx 11.18$$
$$\tan\theta = \frac{5}{10} \quad\Rightarrow\quad \theta \approx 26.6^\circ$$

### Exam tip: The 3D Pythagoras shortcut
For a box with sides $a$, $b$, $c$, the space diagonal is $\sqrt{a^2 + b^2 + c^2}$. Sam's drone is $\sqrt{40^2 + 30^2 + 25^2} = \sqrt{3125} \approx 55.9$ m away.

## Heights from two sightings

A common exam type: you cannot reach the base of a tower, so you measure angles of elevation from two places.

### Worked example: The tower
From point $A$, due south of a tower of height $h$, the angle of elevation of the top is $30^\circ$. From $B$, due east of the tower, it is $45^\circ$. $A$ and $B$ are 100 m apart. Find $h$.

Let the base be $O$. Then:
- In triangle $AOT$: $OA = \dfrac{h}{\tan 30^\circ} = h\sqrt{3}$
- In triangle $BOT$: $OB = \dfrac{h}{\tan 45^\circ} = h$

On the ground, $OA$ points south and $OB$ points east, so angle $AOB = 90^\circ$:
$$OA^2 + OB^2 = AB^2 \quad\Rightarrow\quad 3h^2 + h^2 = 10000 \quad\Rightarrow\quad h = 50 \text{ m}$$

### Watch out: Do not trust the picture
A 3D drawing on flat paper distorts angles. A right angle may not look like one. Decide where right angles are from the facts, not from the look of the sketch.

**Mind map**

```text
3D trigonometry
  Method
    Sketch and label
    Mark right angles
    Find a triangle
    Redraw it flat
  Key facts
    Vertical meets ground at 90°
    Box diagonal √(a² + b² + c²)
  Tools
    Pythagoras
    SOH CAH TOA
    Sine and cosine rules
  Typical problems
    Space diagonals
    Two angles of elevation
    Bearings on the ground
```

### Year 11: Angle between a line and a plane
The angle between a line and a flat surface is the angle between the line and its **shadow** on the surface, if light shines straight down. For $AG$ in the box, the shadow on the floor is $AC$, so the angle is $\theta$ above.

**Angle between two planes:** take a line in each plane, both at right angles to the line where the planes meet, and measure the angle between them.

**Square pyramids.** The top is directly above the centre of the base. The centre is half a diagonal from each corner: for base side $s$ that is $\frac{s}{\sqrt{2}}$. So a pyramid with base 8 and height 6 has sloping edges $\sqrt{32 + 36} = \sqrt{68}$. The angle between a **sloping face** and the base uses the midpoint of a base edge, which is $\frac{s}{2} = 4$ from the centre: $\tan\theta = \frac{6}{4}$, so $\theta \approx 56.3^\circ$.

**Regular tetrahedron** (all edges $a$). The top is above the centre of the equilateral base, which is $\frac{2}{3}$ of the way along a median of length $\frac{\sqrt{3}}{2}a$, so $\frac{a}{\sqrt{3}}$ from each corner. The height is $\sqrt{a^2 - \frac{a^2}{3}} = a\sqrt{\frac{2}{3}}$.

**Face diagonals of a cube** that start at the same corner make $60^\circ$, because joining their ends gives a triangle of three equal face diagonals.

### Year 11: Bearings and the cosine rule
Two hikers leave camp $C$. One walks 3 km on a bearing of $040^\circ$ to $P$. The other walks 5 km on $160^\circ$ to $Q$. The angle $PCQ = 160^\circ - 40^\circ = 120^\circ$. A hill top above $Q$ is 0.4 km high. Find the distance on the ground first:
$$PQ^2 = 3^2 + 5^2 - 2(3)(5)\cos 120^\circ = 49 \quad\Rightarrow\quad PQ = 7 \text{ km}$$
Then the angle of elevation of the hill top from $P$ is $\tan^{-1}\dfrac{0.4}{7} \approx 3.3^\circ$.

**Back bearings.** The bearing of $A$ from $B$ differs from the bearing of $B$ from $A$ by $180^\circ$: if $B$ is on $060^\circ$ from $A$, then $A$ is on $240^\circ$ from $B$.

**Equal elevations.** If the top of a tower has the same elevation $45^\circ$ from $A$ and $B$, then $OA = OB = h$. With angle $AOB = 60^\circ$, triangle $AOB$ is equilateral, so $h = AB$.

**A general formula.** With $A$ south and $B$ east of the tower and elevations $\alpha$ and $\beta$: $OA = h\cot\alpha$ and $OB = h\cot\beta$, so $AB^2 = h^2(\cot^2\alpha + \cot^2\beta)$ and $h = \dfrac{AB}{\sqrt{\cot^2\alpha + \cot^2\beta}}$.

### Year 12: Vectors make 3D easier
In Year 12 you will meet vectors. Many 3D problems become shorter with vectors, because the angle between two lines comes straight from the scalar product.

## Quick summary

- Find right-angled triangles inside the 3D shape and solve them one at a time.
- Any vertical line is at right angles to every line on the ground.
- Box diagonal: $\sqrt{a^2 + b^2 + c^2}$.
- With two angles of elevation, write each ground distance in terms of the height, then link them with Pythagoras or the cosine rule.

## Quiz

Try these questions from the chapter quiz.

1. A tower 20 m tall is seen from 20 m away on flat ground. The angle of elevation of the top is:
   - A. $30^\circ$
   - B. $45^\circ$
   - C. $60^\circ$
   - D. $90^\circ$
   - Answer: B. $45^\circ$. $\tan\theta = \dfrac{20}{20} = 1$.

2. The angle of depression from a cliff top to a boat is $30^\circ$. From the boat, the angle of elevation of the cliff top is:
   - A. $30^\circ$
   - B. $60^\circ$
   - C. $120^\circ$
   - D. $150^\circ$
   - Answer: A. $30^\circ$. They are alternate angles between horizontal lines.

3. A vertical pole stands on flat ground. The angle between the pole and any line on the ground through its base is:
   - A. $45^\circ$
   - B. $90^\circ$
   - C. $180^\circ$
   - D. it depends on the line
   - Answer: B. $90^\circ$. Vertical meets horizontal at a right angle.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/three-dimensional-trigonometry/#quiz