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3D Pythagoras & Trigonometry

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Cambridge IGCSE — Extended Maths

3D Pythagoras & Trigonometry

The Big Idea: any messy 3D shape can be broken down into flat, ordinary 2D right-angled triangles — find the ones hiding inside it, draw them flat on paper, then just use Pythagoras' theorem and SOHCAHTOA like normal.

Summary — What This Chapter Covers

Right-angled triangles are hidden inside 3D shapes (cuboids, prisms, cones, pyramids) — your job is to spot them.
Pythagoras' theorem (a² = b² + c²) finds missing lengths once you've isolated a 2D triangle.
There's a 3D version of Pythagoras — d² = x² + y² + z² — for the straight-line distance between two corners, but you never actually need to memorise it.
SOHCAHTOA finds missing angles or sides once a right-angled triangle is isolated, exactly like in 2D.
The "angle between a line and a plane" is a specific, common exam question — it needs a very particular triangle to be drawn correctly.
Complicated problems are solved by chaining together multiple 2D triangles, one feeding into the next.

1. Pythagoras' Theorem in 3D Shapes

Here's the mindset shift for this whole chapter: 3D geometry is just 2D geometry that's been folded up. Every cone, cuboid, pyramid, and prism you'll ever be asked about is secretly hiding a bunch of ordinary flat triangles inside it. Your entire job is to find one, redraw it flat on your page (ignoring the 3D shape around it), and then solve it exactly the way you've solved 2D triangles for years.

Why this works

A right-angled triangle doesn't care whether it's sitting flat on a page or tilted inside a 3D cone — the relationship a² = b² + c² is exactly the same either way. The 3D-ness of the problem is just packaging. Once you strip a triangle out of the shape, it's a completely normal 2D problem.

Worked Example — The Cone

Problem A cone has base radius 5 cm and perpendicular height 12 cm. Find the slant height, l.

Cone (side view, but we only care about ONE triangle inside it): /| / | l / | 12 cm (perpendicular height) / | /_____| 5 cm (radius) This right-angled triangle is formed by: • the radius (base) • the perpendicular height (vertical) • the slant height (hypotenuse, l)

Step 1 — Isolate the triangle: radius = 5, height = 12, slant = l (hypotenuse).

Step 2 — Apply Pythagoras: l² = 5² + 12² = 25 + 144 = 169

Step 3 — Square root: l = √169 = 13 cm

Core Rule

a² = b² + c²

In any right-angled triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides. This never changes — 2D or 3D.

Practice Question 1.1

A cone has base radius 9 cm and slant height 15 cm. Find the perpendicular height of the cone.

2. The 3D Version of Pythagoras' Theorem

Sometimes you're not given a neat 2D triangle to start with — you're just asked for the distance between two corners of a cuboid, like from one bottom corner to the diagonally opposite top corner. This is where the 3D Pythagoras formula comes in.

3D Distance Formula

d² = x² + y² + z²

d is the straight-line distance between two points. x, y, and z are the distances between those points measured along the three different perpendicular directions (length, width, height).

Important exam tip

You will not be given this formula in the exam. That's not a problem though, because of the next trick...

The "Two 2D Problems" Trick

Every 3D distance question can be solved without memorising the 3D formula at all — you just apply ordinary 2D Pythagoras twice in a row. Here's the logic:

Cuboid: A is bottom-front-left corner, B is top-back-right corner Dimensions: 2 cm x 2 cm x 6 cm B /| / | 2 cm (height) / | /___|____ A | / \ | / 6 cm (length) 2cm\ | / \|/ (base diagonal goes here — find THIS first) Step 1: On the BASE of the cuboid, form a flat right-angled triangle using the two base edges (2 cm and 6 cm). Find the base diagonal using normal Pythagoras. Step 2: NOW use that diagonal as one side of a NEW right-angled triangle, together with the vertical height (2 cm), to find the true 3D distance AB.

Worked Example — Distance Between Two Corners

Problem A cuboid measures 2 cm × 6 cm × 2 cm. Find the distance between opposite corners A and B.

Step 1 — Base triangle: the base is 2 cm by 6 cm. Let a = base diagonal.
a² = 2² + 6² = 4 + 36 = 40 (don't square root yet — you'll need a² again in a moment!)

Step 2 — Vertical triangle: now build a new triangle using the base diagonal (√40) and the vertical height (2 cm).
x² = (√40)² + 2² = 40 + 4 = 44

Step 3 — Solve: x = √44 = 6.63 cm (2 d.p.)

Check using the 3D formula directly: d² = 2² + 6² + 2² = 4 + 36 + 4 = 44 → d = √44 = 6.63 cm ✓ Same answer, two different routes.

Memory trick

Leaving a value "squared" (like a² = 40, not a = 6.32...) instead of rounding it early is one of the highest-value exam habits in this whole topic. Round too early and your final answer drifts off — sometimes by a lot.

Practice Question 2.1

A cuboid has dimensions 3 cm × 4 cm × 12 cm. Find the distance between two diagonally opposite corners, using the "two 2D triangles" method.

3. SOHCAHTOA in 3D

Once you've spotted and isolated a right-angled triangle inside a 3D shape, finding a missing angle works exactly like finding a missing length — draw the triangle flat, label the sides relative to the angle you care about, then apply SOHCAHTOA.

SOHCAHTOA (reminder)

sin θ = O/H cos θ = A/H tan θ = O/A

O = side Opposite the angle, A = side Adjacent to the angle, H = Hypotenuse (longest side, opposite the right angle). Pick whichever ratio uses the two sides you already know.

Worked Example — Angle Inside a Cone

Problem Same cone as before: base radius 5 cm, perpendicular height 12 cm. Find angle θ between the slant edge and the base, to 1 decimal place.

|\ | \ 12 cm | \ (hypotenuse) (opp) | \ |____\ 5 cm θ (adjacent)

Step 1 — Identify sides relative to θ: opposite = 12 (height), adjacent = 5 (radius). We know O and A, so use TOA.

Step 2 — Set up: tan θ = O/A = 12/5 = 2.4

Step 3 — Undo tan with tan⁻¹ (shift + tan on your calculator): θ = tan⁻¹(2.4)

Step 4 — Answer: θ = 67.4° (1 d.p.)

Common mistake

Students often mix up which side is "opposite" and which is "adjacent" once the triangle is tilted inside a 3D drawing. This is exactly why you should always redraw the triangle flat on a blank part of your page first — once it's flat and upright, opposite/adjacent become obvious again, just like in 2D.

Practice Question 3.1

A cone has base radius 6 cm and perpendicular height 8 cm. Find the angle between the slant edge and the base, to 1 d.p.

4. The Angle Between a Line and a Plane

This is the single most commonly-tested — and most commonly botched — skill in this whole chapter. It sounds abstract, but there's a completely reliable method for finding it every single time.

The Method (memorise this order)

Identify the plane named in the question (a plane is a flat face — like the base, or one side of the shape).
Identify (or draw on) the line whose angle to that plane you need.
From the far end of the line, drop a perpendicular height straight down onto the plane — this height must form a right angle with the plane.
Connect that new point back to where the line meets the plane. This creates a right-angled triangle.
The angle you want is the one sitting between the line and the plane inside that triangle — now just use SOHCAHTOA.

Two ways people picture this

The fishing rod trick: imagine the line is a fishing rod. If you dropped a hook straight down (vertically) from the tip of the rod until it hit the plane, where would it land? Connect that landing spot back to where the rod starts — that's your triangle.

The "corner of the room" trick: imagine you're standing inside the 3D shape, in one corner, looking along the floor and up the walls like they're the edges of a room. This helps some people "feel" which face is the plane and which edge is the line.

Cuboid ABCDEFGH — angle between line AF and plane ABCD (the base) E-------F /| /| / | / | A--|----B | | H----|--G | / | / D-------C To find angle between AF and base ABCD: 1. Line is AF (goes from corner A up to opposite-top corner F) 2. Plane is ABCD (the bottom face) 3. Drop a perpendicular from F straight down to the base — it lands exactly on B (since F is directly above B). 4. Connect B back to A. 5. Triangle ABF is now a flat right-angled triangle: - AB is on the plane (adjacent) - BF is the perpendicular height (opposite) - AF is the line itself (hypotenuse) 6. The angle at A (between AF and AB) is the angle you want.

If you're not sure which point to use

Try a different point on the same line and see if it makes the right-angled triangle easier to spot. There's often more than one valid way to draw the same angle — you just need a right-angled triangle that contains it.

5. Multi-Step Problem Solving

The trickiest exam questions won't hand you a ready-made triangle — you'll need to chain together two or more 2D triangles, where the answer from the first becomes an input for the second. The strategy never changes though: keep splitting the 3D shape into flat 2D triangles until you reach the one that actually contains what you're asked to find.

Worked Example — Full Multi-Step Problem

Problem A cuboid box ABCD-EFGH (base ABCD, top EFGH) has dimensions 4 cm × 6 cm × 3 cm (A directly below E, etc.), matching the diagram style below. Find:
(a) the length of the longest pencil that fits in the box (the diagonal AF), and
(b) the angle that this pencil makes with the top face BEFC.

E---------F /| /| / | / | B--|------C | 3cm | | | | | A-------|-D | / 4cm | / |/ |/ (front-bottom edge) 6 cm = depth (front-to-back) Longest diagonal = A to F (opposite corners)

Part (a) — Step 1: Build a flat triangle using the top face edges: BF where B→E is 4 cm and E→F is 6 cm (top rectangle). Using Pythagoras: a² = 4² + 6² = 16 + 36 = 52 (leave un-rooted).

Part (a) — Step 2: Now build the "real" triangle ABF: AB = 3 cm (vertical height), BF = √52 (from step 1), AF = the answer we want (hypotenuse).
b² = 3² + 52 = 9 + 52 = 61

Part (a) — Answer: AF = √61 = 7.81 cm (3 s.f.)

Cross-check with the 3D formula directly: d² = 4² + 6² + 3² = 16+36+9 = 61 → d = √61 = 7.81 cm ✓

Part (b) — Step 1: Plane BEFC is the top face. We need a right-angled triangle containing AF, with a side perpendicular to that top plane. AB (3 cm) is exactly that perpendicular height.

Part (b) — Step 2: Draw triangle ABF flat: AB = 3 cm (opposite the angle we want, at F), BF = √52 cm (adjacent), AF = √61 cm (hypotenuse).

Part (b) — Step 3: All three sides are known, so any ratio works. Using tan: tan c = opposite/adjacent = 3/√52

Part (b) — Answer: c = tan⁻¹(3/√52) = 22.6° (1 d.p.)

Examiner's tip (straight from the mark scheme mindset)

If you're stuck on a complicated 3D diagram in the exam, don't leave it blank. Start calculating any length or angle you can see how to find — even if you're not sure it's the one the question wants. These intermediate results often turn out to be exactly what you needed for the next step, and you'll pick up method marks even if you don't reach the final answer.

Practice Question 5.1

A cuboid has base 5 cm × 12 cm and height 8 cm. Find the length of the space diagonal (corner to opposite corner), using the two-triangle method, then find the angle this diagonal makes with the base.

What to Memorise

Term / Rule	What it means
Pythagoras' theorem	a² = b² + c², where a is the hypotenuse (longest side, opposite the right angle). Works in any right-angled triangle, flat or extracted from 3D.
3D Pythagoras (not given in exam)	d² = x² + y² + z² — direct distance between two 3D points using the three perpendicular directions. Can always be avoided using two 2D triangles instead.
SOHCAHTOA	sin=O/H, cos=A/H, tan=O/A — used to find missing angles or sides once a right-angled triangle is isolated.
Plane	A flat 2D face of a 3D shape (e.g. the base, or one side wall).
Angle between a line and a plane	Formed by dropping a perpendicular from the line down onto the plane, then connecting back — always creates a right-angled triangle.
"Two 2D triangles" method	Any 3D distance problem can be solved with two separate applications of ordinary 2D Pythagoras — no need to memorise the 3D formula.
Leave it squared	When a value like a² will be reused in a second Pythagoras step, don't square-root it early — carry the squared value through to avoid rounding errors.

Concepts Checklist

I can identify a hidden right-angled triangle inside a 3D shape (cone, cuboid, pyramid, prism).
I can redraw a 3D triangle "flat" on paper before solving it.
I can apply Pythagoras' theorem to find a missing length once a triangle is isolated.
I understand the 3D Pythagoras formula (d² = x² + y² + z²) and know it isn't given in the exam.
I can solve a 3D distance problem using two separate 2D Pythagoras steps, without needing the 3D formula.
I know to leave a value as "squared" (not rooted) when I'll need to square it again in the next step.
I can apply SOHCAHTOA to find a missing angle inside an isolated 3D triangle.
I correctly identify opposite/adjacent/hypotenuse even when the triangle is tilted in a 3D drawing.
I can find the angle between a line and a plane using the "drop a perpendicular" method.
I can solve multi-step problems by chaining together more than one 2D triangle.
If I get stuck, I know to start calculating any length/angle I can find rather than leaving the question blank.

Exam Tips & Common Mistakes

Mistake #1 — Rounding too early

If a length like a² = 52 needs to be reused in a second Pythagoras or SOHCAHTOA step, keep it as √52 (or even just 52, unsquared) rather than rounding it to 7.21 first. Rounding early compounds errors and can shift your final answer outside the accepted range.

Mistake #2 — Wrong perpendicular for "line and plane" angles

The height you drop to find the angle between a line and a plane must be perpendicular to that specific plane — not just "some vertical-looking line" in the diagram. Always double-check the drop lands at a right angle to the named plane.

Mistake #3 — Mixing up opposite and adjacent

Once a triangle is drawn "at an angle" as part of a 3D sketch, it's very easy to mislabel which side is opposite vs adjacent to the target angle. Always take the extra 10 seconds to redraw the triangle sitting flat and upright on a spare bit of page before applying SOHCAHTOA.

Mistake #4 — Trying to memorise the 3D formula under pressure

You don't need d² = x² + y² + z² memorised at all. Examiners deliberately don't give it to you because the two-2D-triangles method tests exactly the skill they want to see: can you actually spot triangles inside a 3D shape? Lean on that method — it's more reliable under exam pressure anyway.

Mistake #5 — Leaving a hard question blank

Complicated 3D diagrams intimidate people into leaving them blank. Don't. Start finding whatever lengths and angles you can see how to calculate — even partial working earns method marks, and it very often reveals the missing piece you needed for the final answer.

What examiners are actually testing

This topic is really testing whether you can (1) mentally extract a flat 2D triangle from a 3D object, and (2) apply skills you already know (Pythagoras, SOHCAHTOA) to it without getting confused by the 3D packaging. If you can consistently do step (1), step (2) is nothing new.

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3D Pythagoras & Trigonometry

3D Pythagoras & Trigonometry

Summary — What This Chapter Covers

1. Pythagoras' Theorem in 3D Shapes

Worked Example — The Cone

2. The 3D Version of Pythagoras' Theorem

The "Two 2D Problems" Trick

Worked Example — Distance Between Two Corners

3. SOHCAHTOA in 3D

Worked Example — Angle Inside a Cone

4. The Angle Between a Line and a Plane

Two ways people picture this

5. Multi-Step Problem Solving

Worked Example — Full Multi-Step Problem

What to Memorise

Concepts Checklist

Exam Tips & Common Mistakes

More International Mathematics topics