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11. Applications of Trigonometry
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VCE 11 General 2023

11.04 Trigonometry in 3D

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Pythagoras in 3D

The familiar theorem of Pythagoras applies to right-angled triangles drawn on a plane surface. In this setting, triangles are made of lines in a two-dimensional space. This idea, however, can be extended by considering lines in three-dimensional space.

Consider a box-shaped room. Its walls are perpendicular to the floor and the walls that meet are set at right-angles to one another. A rectangular prism can be used to visualise this.

A rectangular prism with two diagonal lines drawn in it.

We can construct two lines on the 3D diagram, as shown below.

The diagonal across the floor makes a right-angled triangle with the lines formed where the floor meets two of the walls.

The body diagonal from the floor to the ceiling through the room makes a right-angled triangle with the floor diagonal and the line formed by the intersection of two of the walls.

So, we have constructed two right-angled triangles, and the floor diagonal is common to both. If the wall dimensions are known, then Pythagoras' theorem can be used to first find the length of the floor diagonal. Once the floor diagonal is known, its length can be used to calculate the body diagonal, again using Pythagoras' theorem.

A rectangular prism with two diagonal lines, line y and line x. Ask your teacher for more information.

This is done in the working after the following diagram:

Firstly, because of the right-angled triangle on the floor of the room, Pythagoras' theorem states that x^2=d^2+w^2.

Next, because of the right-angled triangle cutting diagonally through the room, Pythagoras' theorem states that y^2=x^2+h^2.

Putting these results together, the body diagonal length can be calculated using the following formula. y^2=d^2+w^2+h^2.

Examples

Example 1

How long is the body diagonal of a box with dimensions 12 \text{ cm},\,19 \text{ cm}, and 7\text{ cm}?

Worked Solution
Create a strategy

Use the formula: y^2=d^2+w^2+h^2.

Apply the idea

A diagram like the one above would have dimensions d=12,\,w=19, and h=7. So:

\displaystyle y^2\displaystyle =\displaystyle 12^2+19^2+7^2Substitute the values
\displaystyle y^2\displaystyle =\displaystyle 554Evaluate
\displaystyle y\displaystyle \approx\displaystyle 23.5 \text{ cm}Take the square root of both sides

Example 2

A square prism has sides of length 9\text{ cm},9\text{ cm}, and 16\text{ cm} as shown.

Square prism with a height of 9 centimetres, length of 16 centimetres, and width of 9 centimetres. Ask your teacher for more information.
a

If the diagonal HF has a length of z\text{ cm}, calculate the exact length of z, leaving your answer in surd form.

Worked Solution
Create a strategy

Use Pythagoras' theorem on the right-angled \triangle FGH.

Apply the idea

HG is the same length as AB. So we can let c=z,\,a=9, and b=16.

\displaystyle c^2\displaystyle =\displaystyle a^2+b^2Use Pythagoras' theorem
\displaystyle z^2\displaystyle =\displaystyle 9^2+16^2Substitute c=z, a=9, and b=16
\displaystyle z^2\displaystyle =\displaystyle 337Evaluate
\displaystyle z\displaystyle =\displaystyle \sqrt{337}\text{ cm}Take the square root of both sides
b

Calculate y to two decimal places.

Worked Solution
Create a strategy

Use Pythagoras' theorem on the right-angled \triangle FHD.

Apply the idea

HD is the same length as BF.So we can let c=y, \, a=z=\sqrt{337}, and b=9.

\displaystyle c^2\displaystyle =\displaystyle a^2+b^2Use Pythagoras' theorem
\displaystyle y^2\displaystyle =\displaystyle \left(\sqrt{337}\right)^2+9^2Substitute c=y, a=\sqrt{337}, and b=9
\displaystyle y^2\displaystyle =\displaystyle 418Evaluate
\displaystyle y\displaystyle =\displaystyle \sqrt{418}Take the square root of both sides
\displaystyle y\displaystyle =\displaystyle 20.45\text{ cm}Evaluate

Example 3

A triangular divider has been placed into a box, as shown in the diagram.

Square prism with a height of 12 centimetres, length of 17 centimetres, and width of 12 centimetres. Ask your teacher for more information.
a

If the length of the base of the divider AC is z\text{ cm}, calculate the exact value of z . leaving your answer in surd form.

Worked Solution
Create a strategy

Use Pythagoras' theorem on the right-angled \triangle ABC .

Apply the idea
\displaystyle c^2\displaystyle =\displaystyle a^2+b^2Use Pythagoras' theorem
\displaystyle z^2\displaystyle =\displaystyle 12^2+17^2Substitute c=z,\,a=12,\,b=17
\displaystyle z^2\displaystyle =\displaystyle 433Evaluate
\displaystyle z\displaystyle =\displaystyle \sqrt{433}\text{ cm}Take the square root of both sides
b

Use the exact value of z from part (a) to calculate the length of the diagonal AD correct to two decimal places.

Worked Solution
Create a strategy

Use Pythagoras' theorem on the right-angled \triangle ACD.

Apply the idea

Let c=AD, \, a=z=\sqrt{433}, and b=12.

\displaystyle c^2\displaystyle =\displaystyle a^2+b^2Use Pythagoras' theorem
\displaystyle AD^2\displaystyle =\displaystyle \left(\sqrt{433}\right)^2+12^2Substitute the values
\displaystyle AD^2\displaystyle =\displaystyle 577Evaluate
\displaystyle AD\displaystyle =\displaystyle 24.02\text{ cm}Take the square root of both sides
c

Calculate the area of the triangular divider correct to two decimal places.

Worked Solution
Create a strategy

Use the formula for calculating the area of a triangle.

Apply the idea

In part (a), we found that the length of the base of the divider, z=\sqrt{433}. We have been given the height of the triangular divider, h=12.

\displaystyle \text{Area}\displaystyle =\displaystyle \dfrac{1}{2}\times\text{Base}\times\text{Height}Write the formula
\displaystyle =\displaystyle \dfrac{1}{2}\times\sqrt{433}\times12Substitute the values
\displaystyle =\displaystyle 124.85 \text{ cm}^2Evaluate
Idea summary

The Pythagorean theorem applies to right-angled triangles on a 2D plane but can be extended to 3D by considering lines in three-dimensional space.

Trigonometry in 3D

Just like Pythagoras' theorem, trigonometry can be applied to right-angled triangles drawn in 3D space.

Examples

Example 4

A square prism has sides of length 3\text{ cm},3\text{ cm}, and 14\text{ cm} as shown.

Square prism with a height of 3 centimetres, length of 14 centimetres, and width of 3 centimetres. Ask your teacher for more information.
a

If the diagonal HF has a length of z\text{ cm}, calculate z to two decimal places.

Worked Solution
Create a strategy

Use Pythagoras' theorem on the right-angled \triangle FGH.

Apply the idea

HG is the same length as AB. So we can let c=z,\,a=3, and b=14.

\displaystyle c^2\displaystyle =\displaystyle a^2+b^2Use Pythagoras' theorem
\displaystyle z^2\displaystyle =\displaystyle 3^2+14^2Substitute the values
\displaystyle z^2\displaystyle =\displaystyle 205Evaluate
\displaystyle z\displaystyle =\displaystyle 14.32\text{ cm}Take the square root of both sides
b

If the size of \angle DFH is \theta \degree, find theta to two decimal places.

Worked Solution
Create a strategy

Determine the sides using the reference angle, then use the appropriate ratio.

Apply the idea

With respect to the angle \theta \degree, the opposite side is 3, and the adjacent side is 14.32, found in part (a). We can use the tangent ratio.

\displaystyle \tan \theta\displaystyle =\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}Use the tangent ratio
\displaystyle \tan \theta\displaystyle =\displaystyle \frac{3}{14.32}Substitute the values
\displaystyle \theta\displaystyle =\displaystyle \tan^{-1}\left(\dfrac{3}{14.32}\right)Apply inverse tangent on both sides
\displaystyle =\displaystyle 11.83 \degreeEvaluate using a calculator
Idea summary

Trigonometry, similar to Pythagoras' theorem, can be applied to right-angled triangles in 3D space. Remember the mnemonic, SOH CAH TOA, as a helpful tool.

Outcomes

U2.AoS4.9

solve practical problems involving right-angled triangles in the dimensions including the use of angles of elevation and depression, Pythagoras’ theorem trigonometric ratios sine, cosine and tangent and the use of three-figure (true) bearings in navigation

U2.AoS4.13

calculate the perimeter, areas, volumes and surface areas of solids (spheres, cylinders, pyramids and prisms and composite objects) in practical situations, including simple uses of Pythagoras’ in three dimensions

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