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3D Trigonometry - Higher

3D Trigonometry - Higher

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Tutor: Bilal

Summary

3D Trigonometry

In a nutshell

33​D trigonometry is very similar to 22​D trigonometry. The key difference is being able to process a 33​D problem into a 22​D one to make it easier to visualise and solve.


Planes and lines in 33​D

In 33D 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 33​D problems to 22​D problems

To convert 33​D trigonometry problems to 22​D ones, you have to be able to identify exactly what angles and sides you need, then draw a 22​D shape that simplifies the question. It is best shown with an example.


Example 1

Give a 22​D representation of the angle between the line AFAF​ and the plane ABGHABGH​ in this cuboid.


Maths; Trigonometry; KS4 Year 10; 3D Trigonometry - Higher


First, identify the line AFAF​ and the plane ABGHABGH​. The line AFAF​ is the diagonal of the cuboid. The plane ABGHABGH​ 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 AFAF​. Another one of the sides will be on the plane ABGHABGH​. To keep this two dimensional, use the line AHAH​ as it is in the same direction as AFAF​. To complete the triangle, the third line has to be FHFH​. Here is an illustration:


Maths; Trigonometry; KS4 Year 10; 3D Trigonometry - Higher


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 ABAB​, BHBH​, and FHFH​ are 7cm7cm, 3cm3cm, 15cm15cm respectively. Find the angle between the line AFAF​ and the plane ABGHABGH to 11​ decimal place.


Maths; Trigonometry; KS4 Year 10; 3D Trigonometry - Higher


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


Maths; Trigonometry; KS4 Year 10; 3D Trigonometry - Higher


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

(AH)2=(AB)2+(BH)2(AH)^2=(AB)^2+(BH)^2​​


Substitute in AB=7cm, BH=3cmAB = 7cm,\, BH =3cm:

(AH)2=72+32(AH)^2=7^2+3^2​​

(AH)2=49+9(AH)^2=49+9​​

(AH)2=58(AH)^2=58​​

AH=58AH=\sqrt{58}


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

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


tan(x)=1558\tan(x)=\frac{15}{\sqrt{58}}​​


x=tan1(1558)=63.0822655...x=\tan^{-1}(\frac{15}{\sqrt{58}})=63.0822655...​​


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

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FAQs - Frequently Asked Questions

How do we transform a 3D trigonometric problem into a 2D one?

How do you solve a 3D trigonometry problem?

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