3d Pythagoras and Trigonometry

What you'll learn

How to spot right-angled triangles — triangles with one 90° angle — inside 3D shapes.
How to use Pythagoras' theorem to find hidden diagonals in cuboids and pyramids.
How to use SOHCAHTOA to find angles in cuboids, prisms and pyramids.
How to find the angle a line makes with a flat face, called a plane.

1. The 2D tools you need first

Before a 3D question becomes manageable, you usually need to turn it into one or two 2D right-angled triangles.

Definition

Pythagoras' theorem

In a right-angled triangle, the hypotenuse is the side opposite the 90° angle. If the two shorter sides are $a$ and $b$ , and the hypotenuse is $c$ , then $a^2+b^2=c^2$ .

A right-angled triangle showing the hypotenuse opposite the 90° angle and the side labels used in Pythagoras' theorem.

Definition

SOHCAHTOA

For an angle $θ\theta$ in a right-angled triangle: $sin⁡θ=oppositehypotenuse\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ , $cos⁡θ=adjacenthypotenuse\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ , and $tan⁡θ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ . The opposite side is across from the angle; the adjacent side touches the angle but is not the hypotenuse.

A right-angled triangle labelled from angle θ to show opposite, adjacent and hypotenuse.

Example

Finding a diagonal on a rectangle

A rectangle has sides 9 cm and 12 cm. Find its diagonal.

The diagonal of a 9 cm by 12 cm rectangle forms a right-angled triangle.

The diagonal splits the rectangle into a right-angled triangle.
Let the diagonal be $d$ . Use Pythagoras.
$d^2=9^2+12^2=225$
Square root to find $d$ .
$d=225=15d=\sqrt{225}=15$
The diagonal is 15 cm.

2. The big idea in 3D

In 3D, the right triangle you need is often hidden. Your job is to draw or imagine the correct flat triangle inside the shape.

Key Idea

3D becomes 2D

Most 3D Pythagoras and trigonometry questions are solved by finding a useful 2D right-angled triangle inside the solid.

A space diagonal is a diagonal that goes through the inside of a 3D shape, such as from one corner of a cuboid to the opposite corner.

Example

Finding a space diagonal in a cuboid

A cuboid has edges 6 cm, 8 cm and 4 cm. Find the diagonal from the bottom-front corner to the opposite top-back corner.

A cuboid showing the base diagonal AC and the space diagonal AG found using Pythagoras twice.

First find the diagonal across the base. Let this be $A C$ .
Use Pythagoras on the base rectangle.
$AC^2=6^2+8^2=100$
So $A C = 10$ cm. Now use the vertical height 4 cm with $A C$ .
Let the space diagonal be $A G$ . Use Pythagoras again.
$AG^2=10^2+4^2=116$
Square root and round.
$AG=116=10.770…AG=\sqrt{116}=10.770\ldots$
The space diagonal is 10.8 cm to 3 significant figures.

Tip

Cuboid shortcut

For a cuboid with three perpendicular edge lengths $l$ , $w$ and $h$ , the space diagonal $d$ satisfies $d^2=l^2+w^2+h^2$ .

3. Finding a missing edge in a cuboid

Sometimes you are given the space diagonal and two edge lengths, and you need to work backwards.

Common Mistake

Adding when you should subtract

If the longest diagonal is already given, do not add all the squares again. Rearrange Pythagoras and subtract the known squared lengths.

Example

Working backwards from a cuboid diagonal

In a cuboid, $A B = 5$ cm, $A E = 6$ cm and the space diagonal $A G = 13$ cm. Find $A D$ .

A cuboid with two known perpendicular edges and the space diagonal, leaving AD as the missing edge.

Use the cuboid diagonal relationship.
$AG^2=AB^2+AD^2+AE^2$
Substitute the known values.
$13^2=5^2+AD^2+6^2$
Rearrange to make $AD^2$ the subject.
$AD^2=13^2-5^2-6^2=108$
Square root and round.
$AD=108=10.392…AD=\sqrt{108}=10.392\ldots$
The missing edge is 10.4 cm to 3 significant figures.

4. Finding angles inside cuboids

Angle notation matters. In angle $E C A$ , the middle letter, $C$ , is the vertex — the point where the angle is measured.

To find an angle, first find the sides of the right-angled triangle, then choose sine, cosine or tangent.

Example

Finding an angle in a cuboid

A cuboid has $A E = 5$ cm, $A D = 6$ cm and $D C = 8$ cm. Find angle $E C A$ .

The cuboid contains triangle EAC, where angle ECA is measured between the space diagonal CE and the base diagonal CA.

Angle $E C A$ is at $C$ . The useful triangle is $E A C$ .
First find the base diagonal $A C$ .
$AC^2=6^2+8^2=100$
So $A C = 10$ cm. In triangle $E A C$ , the side opposite angle $E C A$ is $A E = 5$ cm, and the adjacent side is $A C = 10$ cm.
Use tangent because you have opposite and adjacent.
$tan⁡θ=510\tan\theta=\frac{5}{10}$
Find the angle.
$θ=tan⁡−1(510)=26.565…\theta=\tan^{-1}\left(\frac{5}{10}\right)=26.565\ldots$
Angle $E C A$ is 26.6° to 3 significant figures.

5. The angle a line makes with a plane

A plane is a flat surface. In a 3D question, “plane ABCD” means the flat face containing points A, B, C and D.

Definition

Angle between a line and a plane

The angle between a line and a plane is the angle between the line and its projection on the plane. The projection is like the line’s shadow dropped straight down onto the flat surface.

Example

Angle between a line and the base of a prism

In a prism, $C D = 18$ cm, $A D = 25$ cm, angle $FDC=40∘FDC=40^\circ$ , and $F$ is vertically above $C$ . Find the angle that $A F$ makes with plane $A B C D$ .

The angle between AF and the base plane is shown using its projection AC on plane ABCD.

The projection of $A F$ onto the base plane is $A C$ , so the required angle is angle $F A C$ .
First find the vertical height $C F$ using triangle $F D C$ .
$tan⁡40∘=CF18\tan 40^\circ=\frac{CF}{18}$
Rearrange to find $C F$ .
$CF=18tan⁡40∘=15.103…CF=18\tan 40^\circ=15.103\ldots$
Find the base diagonal $A C$ .
$AC^2=25^2+18^2=949$
So $AC=949=30.805…AC=\sqrt{949}=30.805\ldots$ cm.
Use triangle $A F C$ to find the angle $θ\theta$ .
$tan⁡θ=15.103…30.805…\tan\theta=\frac{15.103\ldots}{30.805\ldots}$
Calculate the angle.
$θ=26.1∘\theta=26.1^\circ$
The line $A F$ makes an angle of 26.1° with the base plane.

6. Square-based pyramids

In a square-based pyramid, the apex is the top point. The vertical height is the perpendicular distance from the apex to the base.

If the apex is vertically above the centre of the square, the centre lies halfway along both diagonals of the square.

Example

Finding an angle in a square-based pyramid

A square-based pyramid has base side 8 cm. The apex $E$ is 12 cm vertically above the centre $O$ of the base. Find angle $E A C$ .

A square-based pyramid where the vertical height EO meets the centre O and angle EAC is found in triangle EAO.

Find the diagonal $A C$ of the square base.
$AC^2=8^2+8^2=128$
So $AC=82AC=8\sqrt{2}$ cm, and $A O$ is half of this.
$AO=42AO=4\sqrt{2}$
Since $O$ lies on $A C$ , angle $E A C$ is the same as angle $E A O$ .
In right-angled triangle $E A O$ , use tangent.
$tan⁡θ=1242\tan\theta=\frac{12}{4\sqrt{2}}$
Calculate the angle.
$θ=64.760…∘\theta=64.760\ldots^\circ$
Angle $E A C$ is 64.8° to 1 decimal place.

For volume, use:

heightV=\frac{1}{3}\times \text{base area}\times \text{vertical height}

Example

Finding the volume from an angle

A square-based pyramid has base side 12 cm. The apex $P$ is vertically above the centre $O$ , and angle $PAC=62∘PAC=62^\circ$ . Find the volume.

Find the base diagonal $A C$ .
$AC=122AC=12\sqrt{2}$
The centre is halfway along the diagonal.
$AO=62AO=6\sqrt{2}$
Use triangle $A O P$ to find the vertical height $P O$ .
$tan⁡62∘=PO62\tan 62^\circ=\frac{PO}{6\sqrt{2}}$
Rearrange.
$PO=62tan⁡62∘=15.956…PO=6\sqrt{2}\tan 62^\circ=15.956\ldots$
The base area is 144 cm². Use the pyramid volume formula.
$V=13×144×15.956…=765.9…V=\frac{1}{3}\times144\times15.956\ldots=765.9\ldots$
The volume is 766 cm³ to 3 significant figures.

Exam technique

In the exam

Redraw the right-angled triangle you are actually using, even if it is hidden inside the 3D shape.
If you need an angle, label opposite, adjacent and hypotenuse from the angle’s position before choosing sine, cosine or tangent.
Keep full calculator values until the final line, then round to the accuracy requested.

Self review

Check yourself

Can you explain why a cuboid space diagonal often needs Pythagoras twice?
When a question asks for the angle a line makes with a plane, can you identify the projection on the plane?
In a square-based pyramid, do you know where the centre of the base is and how to use half the diagonal?

What you'll learn

How to spot right-angled triangles — triangles with one 90° angle — inside 3D shapes.
How to use Pythagoras' theorem to find hidden diagonals in cuboids and pyramids.
How to use SOHCAHTOA to find angles in cuboids, prisms and pyramids.
How to find the angle a line makes with a flat face, called a plane.

1. The 2D tools you need first

Before a 3D question becomes manageable, you usually need to turn it into one or two 2D right-angled triangles.

Definition

Pythagoras' theorem

In a right-angled triangle, the hypotenuse is the side opposite the 90° angle. If the two shorter sides are $a$ and $b$ , and the hypotenuse is $c$ , then $a^2+b^2=c^2$ .

A right-angled triangle showing the hypotenuse opposite the 90° angle and the side labels used in Pythagoras' theorem.

Definition

SOHCAHTOA

A right-angled triangle labelled from angle θ to show opposite, adjacent and hypotenuse.

Example

Finding a diagonal on a rectangle

A rectangle has sides 9 cm and 12 cm. Find its diagonal.

The diagonal of a 9 cm by 12 cm rectangle forms a right-angled triangle.

The diagonal splits the rectangle into a right-angled triangle.
Let the diagonal be $d$ . Use Pythagoras.
$d^2=9^2+12^2=225$
Square root to find $d$ .
$d=225=15d=\sqrt{225}=15$
The diagonal is 15 cm.

2. The big idea in 3D

In 3D, the right triangle you need is often hidden. Your job is to draw or imagine the correct flat triangle inside the shape.

Key Idea

3D becomes 2D

Most 3D Pythagoras and trigonometry questions are solved by finding a useful 2D right-angled triangle inside the solid.

A space diagonal is a diagonal that goes through the inside of a 3D shape, such as from one corner of a cuboid to the opposite corner.

Example

Finding a space diagonal in a cuboid

A cuboid has edges 6 cm, 8 cm and 4 cm. Find the diagonal from the bottom-front corner to the opposite top-back corner.

A cuboid showing the base diagonal AC and the space diagonal AG found using Pythagoras twice.

First find the diagonal across the base. Let this be $A C$ .
Use Pythagoras on the base rectangle.
$AC^2=6^2+8^2=100$
So $A C = 10$ cm. Now use the vertical height 4 cm with $A C$ .
Let the space diagonal be $A G$ . Use Pythagoras again.
$AG^2=10^2+4^2=116$
Square root and round.
$AG=116=10.770…AG=\sqrt{116}=10.770\ldots$
The space diagonal is 10.8 cm to 3 significant figures.

Tip

Cuboid shortcut

For a cuboid with three perpendicular edge lengths $l$ , $w$ and $h$ , the space diagonal $d$ satisfies $d^2=l^2+w^2+h^2$ .

3. Finding a missing edge in a cuboid

Sometimes you are given the space diagonal and two edge lengths, and you need to work backwards.

Common Mistake

Adding when you should subtract

If the longest diagonal is already given, do not add all the squares again. Rearrange Pythagoras and subtract the known squared lengths.

Example

Working backwards from a cuboid diagonal

In a cuboid, $A B = 5$ cm, $A E = 6$ cm and the space diagonal $A G = 13$ cm. Find $A D$ .

A cuboid with two known perpendicular edges and the space diagonal, leaving AD as the missing edge.

Use the cuboid diagonal relationship.
$AG^2=AB^2+AD^2+AE^2$
Substitute the known values.
$13^2=5^2+AD^2+6^2$
Rearrange to make $AD^2$ the subject.
$AD^2=13^2-5^2-6^2=108$
Square root and round.
$AD=108=10.392…AD=\sqrt{108}=10.392\ldots$
The missing edge is 10.4 cm to 3 significant figures.

4. Finding angles inside cuboids

Angle notation matters. In angle $E C A$ , the middle letter, $C$ , is the vertex — the point where the angle is measured.

To find an angle, first find the sides of the right-angled triangle, then choose sine, cosine or tangent.

Example

Finding an angle in a cuboid

A cuboid has $A E = 5$ cm, $A D = 6$ cm and $D C = 8$ cm. Find angle $E C A$ .

The cuboid contains triangle EAC, where angle ECA is measured between the space diagonal CE and the base diagonal CA.

Angle $E C A$ is at $C$ . The useful triangle is $E A C$ .
First find the base diagonal $A C$ .
$AC^2=6^2+8^2=100$
So $A C = 10$ cm. In triangle $E A C$ , the side opposite angle $E C A$ is $A E = 5$ cm, and the adjacent side is $A C = 10$ cm.
Use tangent because you have opposite and adjacent.
$tan⁡θ=510\tan\theta=\frac{5}{10}$
Find the angle.
$θ=tan⁡−1(510)=26.565…\theta=\tan^{-1}\left(\frac{5}{10}\right)=26.565\ldots$
Angle $E C A$ is 26.6° to 3 significant figures.

5. The angle a line makes with a plane

A plane is a flat surface. In a 3D question, “plane ABCD” means the flat face containing points A, B, C and D.

Definition

Angle between a line and a plane

The angle between a line and a plane is the angle between the line and its projection on the plane. The projection is like the line’s shadow dropped straight down onto the flat surface.

Example

Angle between a line and the base of a prism

In a prism, $C D = 18$ cm, $A D = 25$ cm, angle $FDC=40∘FDC=40^\circ$ , and $F$ is vertically above $C$ . Find the angle that $A F$ makes with plane $A B C D$ .

The angle between AF and the base plane is shown using its projection AC on plane ABCD.

The projection of $A F$ onto the base plane is $A C$ , so the required angle is angle $F A C$ .
First find the vertical height $C F$ using triangle $F D C$ .
$tan⁡40∘=CF18\tan 40^\circ=\frac{CF}{18}$
Rearrange to find $C F$ .
$CF=18tan⁡40∘=15.103…CF=18\tan 40^\circ=15.103\ldots$
Find the base diagonal $A C$ .
$AC^2=25^2+18^2=949$
So $AC=949=30.805…AC=\sqrt{949}=30.805\ldots$ cm.
Use triangle $A F C$ to find the angle $θ\theta$ .
$tan⁡θ=15.103…30.805…\tan\theta=\frac{15.103\ldots}{30.805\ldots}$
Calculate the angle.
$θ=26.1∘\theta=26.1^\circ$
The line $A F$ makes an angle of 26.1° with the base plane.

6. Square-based pyramids

In a square-based pyramid, the apex is the top point. The vertical height is the perpendicular distance from the apex to the base.

If the apex is vertically above the centre of the square, the centre lies halfway along both diagonals of the square.

Example

Finding an angle in a square-based pyramid

A square-based pyramid has base side 8 cm. The apex $E$ is 12 cm vertically above the centre $O$ of the base. Find angle $E A C$ .

A square-based pyramid where the vertical height EO meets the centre O and angle EAC is found in triangle EAO.

Find the diagonal $A C$ of the square base.
$AC^2=8^2+8^2=128$
So $AC=82AC=8\sqrt{2}$ cm, and $A O$ is half of this.
$AO=42AO=4\sqrt{2}$
Since $O$ lies on $A C$ , angle $E A C$ is the same as angle $E A O$ .
In right-angled triangle $E A O$ , use tangent.
$tan⁡θ=1242\tan\theta=\frac{12}{4\sqrt{2}}$
Calculate the angle.
$θ=64.760…∘\theta=64.760\ldots^\circ$
Angle $E A C$ is 64.8° to 1 decimal place.

For volume, use:

heightV=\frac{1}{3}\times \text{base area}\times \text{vertical height}

Example

Finding the volume from an angle

A square-based pyramid has base side 12 cm. The apex $P$ is vertically above the centre $O$ , and angle $PAC=62∘PAC=62^\circ$ . Find the volume.

Find the base diagonal $A C$ .
$AC=122AC=12\sqrt{2}$
The centre is halfway along the diagonal.
$AO=62AO=6\sqrt{2}$
Use triangle $A O P$ to find the vertical height $P O$ .
$tan⁡62∘=PO62\tan 62^\circ=\frac{PO}{6\sqrt{2}}$
Rearrange.
$PO=62tan⁡62∘=15.956…PO=6\sqrt{2}\tan 62^\circ=15.956\ldots$
The base area is 144 cm². Use the pyramid volume formula.
$V=13×144×15.956…=765.9…V=\frac{1}{3}\times144\times15.956\ldots=765.9\ldots$
The volume is 766 cm³ to 3 significant figures.

Exam technique

In the exam

Redraw the right-angled triangle you are actually using, even if it is hidden inside the 3D shape.
If you need an angle, label opposite, adjacent and hypotenuse from the angle’s position before choosing sine, cosine or tangent.
Keep full calculator values until the final line, then round to the accuracy requested.

Self review

Check yourself

Can you explain why a cuboid space diagonal often needs Pythagoras twice?
When a question asks for the angle a line makes with a plane, can you identify the projection on the plane?
In a square-based pyramid, do you know where the centre of the base is and how to use half the diagonal?

What you'll learn

1. The 2D tools you need first

2. The big idea in 3D

3. Finding a missing edge in a cuboid

4. Finding angles inside cuboids

5. The angle a line makes with a plane

6. Square-based pyramids

Recap questions

Review 3d Pythagoras and Trigonometry by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

3d Pythagoras and Trigonometry

What you'll learn

1. The 2D tools you need first

2. The big idea in 3D

3. Finding a missing edge in a cuboid

4. Finding angles inside cuboids

5. The angle a line makes with a plane

6. Square-based pyramids

Recap questions

Review 3d Pythagoras and Trigonometry by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

3d Pythagoras and Trigonometry

What you'll learn

1. The 2D tools you need first

2. The big idea in 3D

3. Finding a missing edge in a cuboid

4. Finding angles inside cuboids

5. The angle a line makes with a plane

6. Square-based pyramids

Recap questions

Review 3d Pythagoras and Trigonometry by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

3d Pythagoras and Trigonometry Revision Guide

3d Pythagoras and Trigonometry

What you'll learn

1. The 2D tools you need first

2. The big idea in 3D

3. Finding a missing edge in a cuboid

4. Finding angles inside cuboids

5. The angle a line makes with a plane

6. Square-based pyramids

Recap questions

Review 3d Pythagoras and Trigonometry by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

3d Pythagoras and Trigonometry Revision Guide