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iGCSE (2021 Edition)

20.08 Trigonometry in 3D (Extended)

Lesson

Trigonometry in 3D space works the same way as in 2D space, with the same trigonometric ratios and relations.

The key difference between 3D and 2D space when using trigonometry is that there are a lot more triangles we can find. Using this abundance of triangles, we can use multiple applications of trigonometry to find previously unknown side lengths and angles in a given problem.

We can see how multiple applications of trigonometry work together in 3D space to find missing values in the example below.

Worked example

Nathan is standing $15$15 m due North of a building with a height of $h$h. Erin is standing $20$20 m due East of the building. If the angle of elevation from Nathan to the top of the building is $32^\circ$32°, what is Erin's angle of elevation to the same point?

Think: To find the angle of elevation from Erin to the top of the building, we need to know two side lengths of the triangle formed by Erin and the building. We can find the height of the building using trigonometry in the triangle formed by Nathan and the building.

Do: Using trigonometry in the triangle formed by Nathan and the building, we can find the height $h$h of the building to be:

$h=15\tan32^\circ$h=15tan32°

Using this value, we can then find the missing angle $\theta$θ to be:

$\theta$θ$=$=$\tan^{-1}\left(\frac{h}{20}\right)$tan1(h20)$=$=$\tan^{-1}\left(\frac{15\tan32^\circ}{20}\right)$tan1(15tan32°20)

Rounding to two decimal places, find that:

$\theta=25.11^\circ$θ=25.11°

Reflect: As shown, using trigonometry in 3D space can help us to solve problems by using given values to find unknown side lengths which can then be used to find unknown angles.

It is also important when applying trigonometry to 3D problems is to work backwards from the information we are trying to find, since it will tell us what other values are required. Repeating this until the value required are already known will give us a starting place and method for solving the overall problem.

 

Practice questions

Question 1

A square prism has sides of length $7$7cm, $7$7cm and $17$17cm as shown.

A rectangular prism with the upper face is rectangle ABCD and lower face is rectangle EFGH. The length of edges GF is 17, BF is 7 and AB is 7. The angle formed between the segments DF and FH is labeled theta degrees.

  1. If the diagonal $HF$HF has a length of $z$z cm, calculate $z$z to two decimal places.

  2. If the size of $\angle DFH$DFH is $\theta$θ°, find theta to two decimal places.

Question 2

The following is a right pyramid on a square base with side length $16$16cm. A right pyramid has its apex aligned directly above the centre of its base.

The edge length $VA$VA is $24$24cm.

  1. $W$W is the point on the base directly under $V$V. If $AW$AW has a length of $z$z cm, calculate $z$z to two decimal places.

  2. Now, if the size of $\angle VAW$VAW is $\theta$θ°, find $\theta$θ to two decimal places.

Question 3

A room measures $9$9 metres in length and $5$5 metres in width. The angle of elevation from the bottom left corner to the top right corner of the room is $55$55°.

  1. Find $d$d, the distance from one corner of the floor to the opposite corner of the floor. Leave your answer in surd form.

  2. Find $h$h, the height of the room. Give your answer to 2 decimal places.

  3. Find $x$x, the angle of elevation from the bottom corner of the $9$9m long wall to the opposite top corner of the wall, correct to 2 decimal places.

  4. Find $y$y, the angle of depression from the top corner of the $5$5 m long wall to the opposite bottom corner of the wall, correct to two decimal places.

Outcomes

0580E6.5

Solve simple trigonometrical problems in three dimensions including angle between a line and a plane.

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