Pythagoras and trigonometry in 3D
Pythagoras and trig in 3D (Extended)
- Find a right-angled triangle inside the solid, then use Pythagoras or trig on it.
- A common task: the angle between a space diagonal and the base.
Practice
A box has a base 6 cm by 8 cm. Find the length of the base diagonal (cm).
√(6² + 8²) = √100 = 10 cm.
Practice
In a 3-D problem you look for a right-angled triangle inside the solid to work with.
The whole 3-D method is to reduce it to a flat right-angled triangle.
Worked example
- A box has base $6\ \text{cm} \times 8\ \text{cm}$, height $5\ \text{cm}$.
- Base diagonal $= \sqrt{6^2 + 8^2} = \sqrt{100} = 10\ \text{cm}$.
- This diagonal and the height make a right-angled triangle: $\tan\theta = \tfrac{5}{10} = 0.5$, so $\theta = 26.6^{\circ}$.
Practice
The base diagonal is 10 cm and the box height is 5 cm. The angle to the base has tan θ = 5/10. Find θ (degrees, 1 dp).
θ = tan⁻¹(0.5) = 26.6°.
You've got it
Key idea
- in 3D, spot a right-angled triangle inside the solid, then use Pythagoras/trig
- often find a base diagonal first, then bring in the height
- box $6\times8\times5$: base diagonal $10$, diagonal-to-base angle $26.6^{\circ}$