# 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:*

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

$\tan(x)=\frac{15}{\sqrt{58}}$

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

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