Problem Solving: Pythagoras & Trigonometry in 3D

 

Using Pythagoras' Theorem in 3D Shapes

In 3D geometry, many problems can be simplified by spotting right-angled triangles inside the shape. If two sides of a right-angled triangle are known, the third can be found using:

$$a^2 + b^2 = c^2$$

This works well in 3D when you identify a flat triangle within the shape.

If you're trying to find the diagonal between two corners of a cuboid, you can:

 

 

First use Pythagoras in the base to find the diagonal across the bottom face

Then use that diagonal with the vertical side to find the full diagonal (space diagonal)

 

 

 

 

Alternatively, use the 3D version:

$$d^2 = x^2 + y^2 + z^2$$

Where $d$ is the diagonal from one corner of the cuboid to the opposite corner.

 

 
Using Trigonometry (SOHCAHTOA) in 3D

You can also apply SOHCAHTOA in 3D as long as you're working with right-angled triangles. It's useful when finding angles between lines and planes, or between two lines.

Steps:

1. Identify or form a right-angled triangle
2. Label the triangle with the sides relative to the angle: Opposite, Adjacent, Hypotenuse
3. Use one of the trig ratios:
   • $\sin \theta = \frac{O}{H}$
   • $\cos \theta = \frac{A}{H}$
   • $\tan \theta = \frac{O}{A}$
4. Use the inverse function (e.g. $\tan^{-1}$) to find the angle if needed

 

For example:

 

 

Finding Angles Between Lines and Planes

To calculate the angle between a sloping line and a flat surface (plane):

• Identify a right-angled triangle where one side lies flat on the plane and one is vertical
• The sloped line is the hypotenuse
• Use SOHCAHTOA to calculate the angle

Think of the sloped line as a fishing rod and the vertical line as the fishing line dropped to the surface

 
Example: Longest Diagonal in a Cuboid

A box measures $3 cm \times 4 cm \times 6 cm$. What is the length of the longest object that can fit inside?

 

 

Method 1: Using Two Right-Angled Triangles

 

 

Base diagonal ($BF$):

 

 

$$BF = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$$

 

Full diagonal ($AF$):

 

 

$$AF = \sqrt{3^2 + (\sqrt{52})^2} = \sqrt{9 + 52} = \sqrt{61} \approx 7.81 \text{ cm}$$

 

Method 2: 3D Pythagoras

$$d = \sqrt{3^2 + 4^2 + 6^2} = \sqrt{9 + 16 + 36} = \sqrt{61} \approx 7.81 \text{ cm}$$

Final Answer: The longest pencil is 7.81 cm (3 s.f.)

 
Example: Angle Between Line and Plane

Using the same cuboid:

 

 

Find the angle between the diagonal ($AF$) and the base face (plane $BEFC$)

Use triangle $ABF$:

 

 

• Opposite side: 3 cm
• Adjacent (base diagonal $BF$): $\sqrt{52}$

Use tangent:

$$\tan r\degree = \frac{3}{\sqrt{52}} \Rightarrow r\degree = \tan^{-1} \left(\frac{3}{\sqrt{52}}\right) \approx 22.6\degree$$

Final Answer: $AF$ makes an angle of $22.6\degree$ with the base of the box (1 d.p.)