3D Trigonometry

In a nutshell

$$3$$​D trigonometry is very similar to $$2$$​D trigonometry. The key difference is being able to process a $$3$$​D problem into a $$2$$​D one to make it easier to visualise and solve.

Planes and lines in $$3$$​D

In $$3$$D trigonometry, you need to know what a plane is and be able to find angles between lines and planes.

Definition: A plane is a flat, two dimensional surface.

Converting $$3$$​D problems to $$2$$​D problems

To convert $$3$$​D trigonometry problems to $$2$$​D ones, you have to be able to identify exactly what angles and sides you need, then draw a $$2$$​D shape that simplifies the question. It is best shown with an example.

Example 1

Give a $$2$$​D representation of the angle between the line $$AF$$​ and the plane $$ABGH$$​ in this cuboid.

First, identify the line $$AF$$​ and the plane $$ABGH$$​. The line $$AF$$​ is the diagonal of the cuboid. The plane $$ABGH$$​ is the bottom rectangle of the cuboid. To identify the angle between the line and place, you need to draw a triangle.

One of the sides of this triangle is $$AF$$​. Another one of the sides will be on the plane $$ABGH$$​. To keep this two dimensional, use the line $$AH$$​ as it is in the same direction as $$AF$$​. To complete the triangle, the third line has to be $$FH$$​. Here is an illustration:

The transformation from a 3D drawing to a 2D one simplifies the problem and makes it much easier to solve.

Solving 3D trigonometry problems

To solve a 3D trigonometry problem, you first have to convert it to a 2D one and then solve like normal.

Example 2

In the cuboid from the previous example, the sides $$AB$$​, $$BH$$​, and $$FH$$​ are $$7cm$$, $$3cm$$, $$15cm$$ respectively. Find the angle between the line $$AF$$​ and the plane $$ABGH$$​ to $$1$$​ decimal place.

The above example already transformed the problem to a 2 dimensional one. Now, you can use SOHCAHTOA to find the missing angle.

However, there is not enough information to find the angle - at least 2 sides are needed. Therefore, use Pythagoras to find the side $$AH$$​ by focusing on the triangle $$ABH$$​:

​$$(AH)^2=(AB)^2+(BH)^2$$​​

​Substitute in $$AB = 7cm,\, BH =3cm$$:

​$$(AH)^2=7^2+3^2$$​​

​$$(AH)^2=49+9$$​​

​$$(AH)^2=58$$​​

​$$AH=\sqrt{58}$$

​

Now, use trigonometric ratios to find the desired angle - call it $$x$$ for simplicity:

​$$\tan(x)=\frac{opp}{adj}=\frac{FH}{AH}$$​​

​$$\tan(x)=\frac{15}{\sqrt{58}}$$​​

​$$x=\tan^{-1}(\frac{15}{\sqrt{58}})=63.0822655...$$​​

​$$\underline{x=63.1 \degree \ (1 \ d.p.)}$$