Victorian Curriculum Year 10A - 2020 Edition
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7.10 Trigonometry in 3D
Lesson

Much like how we can use Pythagoras' theorem in 3D space, we can also use trigonometry if we have right-angled triangles and the required starting information.

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

The diagram shows a rectangular prism.

  1. Use Pythagoras' theorem to find the length $x$x.

  2. Hence find the length of the prism diagonal $y$y.

  3. Find the angle $\theta$θ.

    Round your answer to the nearest degree.

Question 2

A tree stands at the corner of a square playing field. Each side of the square is $120$120 m long. At the centre of the field the tree subtends an angle of $22^\circ$22°.

  1. Find the distance from the tree to the centre of the field. Round your answer to two decimal places.

  2. Find the height of the tree. Round your answer to two decimal places.

  3. Find the size of the angle subtended by the tree at the corner of the field next to the tree. Round your answer to two decimal places.

  4. Find the size of the angle subtended by the tree at the corner of the field opposite the tree. Round your answer to two decimal places.

Question 3

Two straight paths to the top of a cliff are inclined at angles of $24^\circ$24° and $21^\circ$21° to the horizontal.

  1. If path $A$A is $115$115 m long, find the height $h$h of the cliff.

    Round your answer to the nearest metre.

  2. Find the length $x$x of path $B$B.

    Round your answer to the nearest metre.

  3. If the paths meet at $46^\circ$46° at the base of the cliff, find their distance apart, $y$y, at the top of the cliff.

    Round your answer to the nearest metre.

Outcomes

VCMMG370 (10a)

Apply Pythagoras’ theorem and trigonometry to solving three-dimensional problems in right-angled triangles.

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