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E6.6 Pythagoras’ theorem and trigonometry

Syllabus
0580–2028–2029
Topic
E6.6
Level
Extended

Solve three-dimensional Pythagoras and trigonometry problems

A three-dimensional problem is solved by locating one or more two-dimensional right triangles inside the solid. A line's angle with a plane is the angle between the line and its perpendicular projection onto that plane.

  1. Mark the target line and the relevant plane. 2. Drop the line's endpoint perpendicularly to the plane; join the other endpoint to this foot to form the projection. 3. Find the projection length within the plane, often by Pythagoras. 4. Use the projection, perpendicular height and target line as one right triangle. 5. Apply Pythagoras or the appropriate trig ratio and keep intermediate values unrounded.

d=\sqrt{l^2+w^2+h^2}

A cuboid is 2020 cm long, 5.55.5 cm wide and has volume 495495 cm3^3, so its height is 495/(20×5.5)=4.5495/(20\times5.5)=4.5 cm. The projection of the space diagonal onto the base is 202+5.52\sqrt{20^2+5.5^2} cm. Therefore tan⁡θ=4.5/202+5.52\tan\theta=4.5/\sqrt{20^2+5.5^2}, giving the line-base angle θ=12.2∘\theta=12.2^\circ.

Do not use an arbitrary visible edge as the projection: the projection must lie in the named plane and connect to the perpendicular foot. The requested line-plane angle is the smaller angle in this right triangle, not the complementary angle with the vertical. Use consistent linear units and round only the final result.