Reduce 3D to a 2D triangle: Almost every 3D angle problem becomes a 2D right-angled triangle once you draw the right lines. Find that triangle inside the solid, label its sides, and use SOH-CAH-TOA.
Inside a cuboid: switch between the base diagonal (across the floor), the space diagonal (corner to opposite corner), and the angle the space diagonal makes with the base — the right triangle is base-diagonal, height, space-diagonal.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Draw the triangle separately: Redraw the key triangle flat on its own, with the lengths you know — it's far easier to see the trig than on the 3D picture.
Find the right lengths first: You often need a 3D distance (e.g. a base diagonal) before you can use the triangle — compute that, then do the angle.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Drop a perpendicular to the plane: The angle between a line and a plane is the angle between the line and its projection (shadow) on the plane. Drop a perpendicular from the line's top point to the plane to form the right-angled triangle.
Redraw the 3D angle as a flat right triangle: the height is the opposite side, the base distance the adjacent. Tap tan θ to see the angle is opposite ÷ adjacent.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
IB-style question — diagonal and the base
A vertical pole of height 5 stands at one corner of a square base; a wire runs from its top to the opposite corner, 12 away along the base. Find the angle the wire makes with the base.
Step by step
- The angle a line makes with a plane sits in a right-angled triangle. The 'opposite' is the side facing the angle (here the vertical pole); the 'adjacent' is the side along the base. We know both of those legs and want the angle, so use tan — it links the angle to opposite ÷ adjacent. (sin and cos would need the wire, the hypotenuse, which we aren't told.)
- Substitute the question's numbers: opposite = pole height 5, adjacent = base distance 12.
- Take the inverse tangent to find the angle.
Final answer
About 22.6°.
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
This question, drawn to scale. Opposite the angle θ is the pole (5); next to it is the base (12); the wire is the hypotenuse. We have the two highlighted legs and want θ — so use tan (opposite ÷ adjacent): tan θ = 5/12. sin or cos would need the wire's length, which the question never gives.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
The base distance may need Pythagoras: If the 'adjacent' is a base diagonal, find it first with the 2D (or 3D) distance, then do the angle.
Practice with real exam questions
Answer exam-style questions and get AI feedback that shows you exactly what examiners want to see in a full-marks response.
Lengths first, then the angle: In a box or pyramid, you often need a face or base diagonal before the angle triangle is complete. Use Pythagoras for the diagonal, then SOH-CAH-TOA.
A face of the cube is a 4 × 4 square. Put one corner at A(0, 0) and the opposite at B(4, 4): the face diagonal AB is √(4² + 4²) = √32 = 4√2 ≈ 5.66. The small square marks the right angle where the two sides meet.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
IB-style question — face diagonal of a cube
Find the length of a face diagonal of a cube of side 4.
Step by step
- A face is a rectangle; its diagonal comes from Pythagoras on the two side lengths l and w (the straight line across the corners of the face).
- Substitute the cube's face dimensions: l = 4 and w = 4.
- Simplify the surd.
Final answer
Face diagonal = 4√2 ≈ 5.66.
Two-step problems are normal: 3D questions usually chain: a length (Pythagoras) feeds an angle (trig). Show both steps.
A diameter makes a right angle: The angle in a semicircle is 90°: any point on a circle joined to the ends of a diameter forms a right-angled triangle. This is the trick behind many cylinder-in-sphere problems.
Move P around the circle — joined to the two ends of a diameter, the angle ∠APB is always 90°. That right angle is what unlocks cylinder-in-sphere problems.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
IB-style question — cylinder in a sphere
A cylinder is inscribed in a sphere of radius r so a base diameter and a sphere diameter meet at angle θ. Show that the cylinder's base radius is r cos θ.
Step by step
- A sphere diameter always meets the edge at a right angle (the angle in a semicircle), so the base diameter, the sphere diameter and the cylinder's side make a right-angled triangle — with the sphere diameter as the hypotenuse. In a right triangle, cosine = adjacent ÷ hypotenuse.
- Name the radii: r is the sphere's radius (given) and R is the cylinder's base radius (what we want). A diameter is TWICE the radius — so the sphere's diameter is 2r and the base's diameter is 2R.
- The side next to θ (adjacent) is the base diameter 2R; the hypotenuse is the sphere diameter 2r.
- Top and bottom both have a 2, so the 2s cancel — leaving just the radii.
- Rearrange for the base radius R.
Final answer
So R = r cos θ — exactly the audited exam relationship.
The cylinder (grey) inside the sphere. 2R is the base diameter (straight across the bottom) and 2r is a sphere diameter — the slanted line running through the centre O. They meet at θ. The shaded right triangle is exactly the flat one below: 2R adjacent, 2r hypotenuse.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
The same right triangle, flat: r = the sphere's radius, R = the base radius. Each diameter is just twice its radius, so the base across is 2R (adjacent) and the sphere across is 2r (hypotenuse). cos θ = 2R ÷ 2r; the two 2s cancel → R ÷ r, so R = r cos θ.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Look for diameters: If a diameter appears, suspect a hidden right angle — it usually unlocks the triangle.