Welcome to 3D Trigonometry!
3D Trigonometry combines Pythagoras' theorem and trigonometric ratios to solve problems in three dimensions — finding lengths and angles in cuboids, pyramids, prisms and real-world 3D situations.
Space diagonal: d = √(l² + w² + h²) | Always identify the right-angle triangle first!
Key Skill: The secret to 3D trig is always reducing the problem to a 2D right-angled triangle. Draw it, label it, then use SOHCAHTOA or Pythagoras.
3D Pythagoras
Space diagonals in cuboids and 3D distance formula
Line-Plane Angle
Angle between a line and its projection onto a plane
Dihedral Angle
Angle between two planes meeting along a line
Worked Examples
6 fully solved exam-style questions
3D Cuboid Visualiser
Adjust dimensions and see the space diagonal
Practice 25q
Comprehensive self-marking exercise
Learn 1 — 3D Pythagoras
Space Diagonal of a Cuboid
A cuboid with dimensions l × w × h has a space diagonal (the longest internal diagonal) given by a two-step process:
Step 1: Find the base diagonal: dbase = √(l² + w²)
Step 2: Apply Pythagoras again in the vertical plane: dspace = √(dbase² + h²) = √(l² + w² + h²)
Space diagonal = √(l² + w² + h²)
It is important in exams to show both steps separately — don't just quote the formula. Show the base diagonal first, then find the full diagonal.
Worked Example — Cuboid 3 × 4 × 12
Base diagonal: d = √(3² + 4²) = √(9 + 16) = √25 = 5
Space diagonal: d = √(5² + 12²) = √(25 + 144) = √169 = 13
This is a Pythagorean triple hidden in 3D: (3,4,5) then (5,12,13). Recognising these saves time!
3D Distance Formula
The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in 3D space:
d = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²)
This is simply Pythagoras applied twice — once across the base, once vertically. The x and y differences give the base distance; z gives the height.
Pyramid — Slant Edge from Apex to Base Corner
For a square pyramid with square base a × a and height h (apex directly above the centre):
Find half-diagonal of base: d = √(a² + a²)/2 = a√2/2
Slant edge (apex V to corner A): VA = √(d² + h²) = √(a²/2 + h²)
Triangular Prism Diagonal
For a triangular prism with right-triangle cross-section (legs p and q) and length L:
Longest diagonal: connects a corner of one triangular face to the opposite corner of the other face.
Find the hypotenuse of the triangular face first, then apply Pythagoras with the length L.
Watch out: Always identify the specific right-angle triangle you are using. Label the vertices on a sketch. It is easy to mix up the base, height and slant in 3D.
Learn 2 — Angle Between a Line and a Plane
The angle between a line and a plane is defined as the angle between the line and its orthogonal projection onto the plane — this is always the smallest possible angle between the line and any line in the plane.
Method
Step 1: Find the foot of the perpendicular from the far end of the line down to the plane. Call this point F.
Step 2: The projection of the line onto the plane runs from the base of the line (where it meets the plane) to F.
Step 3: You now have a right-angled triangle: (original line) as hypotenuse, (projection) as base, (perpendicular height) as vertical side.
Step 4: Use SOHCAHTOA: tan(angle) = opposite/adjacent = height/projection length.
Cuboid ABCDEFGH
Label the cuboid so that ABCD is the base and E is above A, F above B, G above C, H above D. AB = l, BC = w, CG = h.
Angle between AG and base ABCD:
• The projection of G onto the base is C (foot of perpendicular from G)
• The projection of AG onto the base is therefore AC
• Find AC: AC = √(AB² + BC²) = √(l² + w²)
• In right triangle AGC: tan(∠GAC) = GC/AC = h/√(l² + w²)
• So ∠GAC = tan⁻¹(h/√(l² + w²))
Example: Cuboid 6 × 8 × 5. AC = √(36+64) = 10. tan(∠GAC) = 5/10 = 0.5. ∠GAC = tan⁻¹(0.5) ≈ 26.6°
Square Pyramid — Angle Between Slant Edge and Base
Square pyramid VABCD, apex V directly above centre O, base side a, height h.
Angle between VA and base ABCD:
• Foot of perpendicular from V is centre O
• Projection of VA onto base is OA (half-diagonal of base)
• OA = a√2/2
• tan(∠VAO) = VO/OA = h/(a√2/2) = 2h/(a√2)
• Note: ∠VAO is the angle between slant edge VA and the base
Always draw a separate 2D right-angle triangle showing the three lengths. Label which is opposite, adjacent, hypotenuse relative to the angle you want.
Angle of Elevation in 3D Context
In real-world problems (hills, buildings, antennas), the angle of elevation is the angle between a horizontal line and the line of sight upward. In 3D you must first find the horizontal distance using Pythagoras before finding the angle.
Horizontal distance between two points = √((Δx)² + (Δy)²)
Then: tan(elevation angle) = vertical height / horizontal distance
Learn 3 — Angle Between Two Planes (Dihedral Angle)
The dihedral angle is the angle between two planes along their line of intersection.
Method
Step 1: Identify the line of intersection of the two planes.
Step 2: In plane 1, draw a line perpendicular to the line of intersection (from any convenient point).
Step 3: In plane 2, draw a line perpendicular to the same line of intersection from the same point.
Step 4: The dihedral angle is the angle between these two perpendicular lines.
Square Pyramid VABCD
Square base ABCD with side a, apex V directly above centre O, height h. Find the dihedral angle between face VAB and base ABCD.
Line of intersection: AB (the shared edge between the two planes)
Midpoint of AB: Call it M
In base ABCD: OM ⊥ AB (by symmetry, O is centre, M is midpoint of AB). OM = a/2 (apothem of base)
In face VAB: VM ⊥ AB (V is above O, so VM is the slant height). VM = √(h² + (a/2)²)
Dihedral angle ∠VMO: tan(angle) = VM/OM ... wait — we need the angle at M.
At M: angle between OM (in base) and VM (in face) = dihedral angle.
tan(dihedral angle) = VO/OM = h/(a/2) = 2h/a
Example: Base 8cm, height 6cm. Apothem OM = 4cm. tan(dihedral) = 6/4 = 1.5. Dihedral angle = tan⁻¹(1.5) ≈ 56.3°
Wedge / Triangular Prism Dihedral Angle
For a triangular prism with a right-angle at one edge, the dihedral angle between the sloping face and the base is found by identifying the foot of the perpendicular from the sloping edge to the base edge.
Combined 3D Problems
Exam questions often combine angles of elevation with 3D settings — a person at point P looks up to point Q on a hill. You must:
- Find the horizontal distance between P and Q using 2D Pythagoras
- Use tan(angle) = vertical height / horizontal distance
Ship problems: "A ship is 5km North and 3km East. A lighthouse is 0.1km tall. Find the angle of elevation."
Horizontal distance = √(5² + 3²) = √34 km. tan(angle) = 0.1/√34. Angle = tan⁻¹(0.1/√34) ≈ 0.98°
Exam Tip: In any 3D problem, always explicitly state: "I will find the right-angled triangle in the plane..." This shows the examiner your method clearly, even if your arithmetic goes slightly wrong.
Example 1 — Space Diagonal of 2 × 3 × 6 Cuboid
Given: Cuboid with l=2, w=3, h=6
Base diagonal: d = √(2² + 3²) = √(4 + 9) = √13
Space diagonal: d = √((√13)² + 6²) = √(13 + 36) = √49 = 7
Final answer is exactly 7 — a clean Pythagorean triple (√13, 6, 7) in disguise!
Example 2 — 3D Distance Between (1,2,3) and (4,6,3)
Given: P(1,2,3) and Q(4,6,3)
Δx = 4−1 = 3, Δy = 6−2 = 4, Δz = 3−3 = 0
d = √(3² + 4² + 0²) = √(9 + 16 + 0) = √25 = 5
The z-coordinates are equal, so this reduces to a 2D distance problem.
Example 3 — Angle Between Diagonal AG and Base in 5×5×10 Cuboid
Given: Cuboid ABCDEFGH with AB=5, BC=5, CG=10
Base diagonal AC: AC = √(5² + 5²) = √50 = 5√2
Right triangle AGC: GC = 10 (vertical), AC = 5√2 (horizontal)
tan(∠GAC) = GC/AC = 10/(5√2) = 2/√2 = √2
∠GAC = tan⁻¹(√2) ≈ 54.7°
Example 4 — Angle Between Slant Edge and Base in Square Pyramid
Given: Square pyramid, base 6×6, vertical height 4cm
Half-diagonal of base: OA = (6√2)/2 = 3√2 cm
Slant edge VA: VA = √((3√2)² + 4²) = √(18 + 16) = √34 cm
tan(∠VAO) = VO/OA = 4/(3√2)
∠VAO = tan⁻¹(4/(3√2)) = tan⁻¹(4/4.243) ≈ tan⁻¹(0.9428) ≈ 43.3°
Example 5 — Dihedral Angle, Square Pyramid Base 8cm Height 6cm
Given: Square pyramid VABCD, base 8cm, height 6cm
Find dihedral angle between triangular face VAB and base ABCD.
M = midpoint of AB: OM = 8/2 = 4cm (apothem of base)
VM (slant height): VM = √(6² + 4²) = √(36+16) = √52 = 2√13 cm
At M: right angle triangle with VO=6 vertical, OM=4 base. tan(dihedral) = 6/4 = 1.5
Dihedral angle = tan⁻¹(1.5) ≈ 56.3°
Example 6 — 3D Bearing and Angle of Elevation
Given: Observer at O. Point A is 12m North, 5m East of O, and 3m above ground. Find the angle of elevation of A from O.
Horizontal distance: d = √(12² + 5²) = √(144 + 25) = √169 = 13m
tan(elevation angle) = 3/13
Angle of elevation = tan⁻¹(3/13) ≈ 13.0°
Common Mistakes in 3D Trigonometry
Mistake 1: Not showing step-by-step method for space diagonal
Students often just write d = √(l²+w²+h²) without working. Examiners want to see the base diagonal calculated first, then the space diagonal. Show both steps for full marks.
Mistake 2: Measuring angle between line and PERPENDICULAR to plane
The angle between a line and a plane is measured from the plane (the projection), NOT from the perpendicular. If you use the perpendicular, you get the complement (90° − answer). Always use the projection as the base of your right triangle.
Mistake 3: Confusing slant height with slant edge in a pyramid
• Slant height: from apex V to midpoint of a base edge (e.g. VM where M is midpoint of AB) — used for dihedral angles
• Slant edge: from apex V to a base corner (e.g. VA) — used for edge length
These are different lengths! The slant height is shorter than the slant edge.
Mistake 4: Applying Pythagoras only once in 3D
Finding the space diagonal requires Pythagoras TWICE. Step 1: base diagonal. Step 2: space diagonal. Students who apply it once get the base diagonal and stop — this is only correct if h=0.
Mistake 5: Not drawing a 2D cross-section
Every 3D trig problem can be solved by finding the correct 2D right-angle triangle inside the shape. Sketch that triangle separately, label all three sides, then apply SOHCAHTOA or Pythagoras. This prevents confusion about which sides to use.
Exam Strategy: When you see a 3D shape, ask yourself: "Which right-angle triangle contains the length or angle I need?" Draw that triangle as a 2D sketch. Label the vertices from the original 3D shape. Then solve the 2D problem.