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.
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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.
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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.
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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°.
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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.
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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.
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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.
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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.
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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.
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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.
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Look for diameters: If a diameter appears, suspect a hidden right angle — it usually unlocks the triangle.