3D Pythagoras & Trigonometry (AQA GCSE Further Maths)

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3D Pythagoras & Trigonometry

What skills will I need to solve 3D problems involving Pythagoras & Trigonometry?

  • Pythagoras’ Theorem helps us find missing side lengths of a right-angled triangle; a squared equals b squared plus c squared

    • It is also frequently used for finding the distance/length of a line

  • SOHCAHTOA is an acronym for the three trigonometric ratios that connect angles (θ) and sides (opposite, hypotenuse and adjacent) in a right-angled triangle

    • sine – SOH – S in space theta equals O over H

    • cosine – CAH – C os space theta equals A over H

    • tangent – TOA – T an space theta equals O over A

  • The hardest problems may involve sine rule, cosine rule and/or area of a triangle

    • sine rule - fraction numerator a over denominator sin space A end fraction equals fraction numerator b over denominator sin space B end fraction equals fraction numerator c over denominator sin space C end fraction

    • cosine rule - a squared equals b squared plus c squared minus 2 b c cos space A

    • area of a triangle - 1 half a b sin space C

  • Other skills involving angles may also be involved

    • e.g. bearings, angles of elevation and depression, angles in parallel and perpendicular lines

How does Pythagoras work in 3D?

  • 3D shapes can often be broken down into several 2D shapes

    • For example nets and surface area

  • With Pythagoras’ Theorem problems you will be specifically looking for right‑angled triangles

    • The right-angled triangles you need will have two known sides and one unknown side

3DPythagTrig Notes fig3 (1), downloadable IGCSE & GCSE Maths revision notes
3DPythagTrig Notes fig3 (2), downloadable IGCSE & GCSE Maths revision notes
3DPythagTrig Notes fig3 (3), downloadable IGCSE & GCSE Maths revision notes
  • There is a 3D version of the Pythagoras’ Theorem formula

    • d squared equals x squared plus y squared plus z squared

  • However it is usually far easier to see a problem by splitting it into two or more 2D problems

3DPythagTrig Notes fig4, downloadable IGCSE & GCSE Maths revision notes

How does trigonometry work in 3D?

  • Again look for a combination of triangles that would lead to the missing angle or side

    • Ideally look for right-angled triangles so SOHCAHTOA can be used

    • Failing that, look for 'pairs' to use sine rule and 'two sides and the angle between them' for cosine rule

  • The angle you are working with can be awkward in 3D

    • The angle between a line and a plane is not obvious

    • If unsure, put a point on the line and draw a new line to the plane

      This should create a right-angled triangle

3DPythagTrig Notes fig5, downloadable IGCSE & GCSE Maths revision notes
3DPythagTrig Notes fig6, downloadable IGCSE & GCSE Maths revision notes
  • Once you have your 2D triangle(s) you can begin to solve problems

3DPythagTrig Notes fig7 (1), downloadable IGCSE & GCSE Maths revision notes
3DPythagTrig Notes fig7 (2), downloadable IGCSE & GCSE Maths revision notes
3DPythagTrig Notes fig7 (3), downloadable IGCSE & GCSE Maths revision notes

Examiner Tips and Tricks

  • Add lines/triangles/etc to any given diagram to help you see the problem

    • If no diagram is given, sketch your own!

  • Draw any 2D triangles separately as a 3D diagram can get hard to follow

Worked Example

The diagram shows a vertical tree, PQ.
The angle of elevation of the top of the tree from the point R on the ground, x m due east of the tree, is 32°.
The point S lies on the ground, 2x m due north of the tree.
The distance between the points R and S is 45 m.

Find the height of the tree, giving your answer to the nearest 10 cm.

picture-4

Draw triangle RQS as a 2D diagram

3d-pythag-cross1

Use Pythagoras' theorem to find x 

 table row cell x squared plus open parentheses 2 x close parentheses squared end cell equals cell 45 squared end cell row cell x squared plus 4 x squared end cell equals 2025 row cell 5 x squared end cell equals 2025 row cell x squared end cell equals 405 row x equals cell square root of 405 end cell row x equals cell 9 square root of 5 end cell end table 

Draw triangle RQP and include 9√5 on the diagram

3d-pythag-cross2

Use SOHCAHTOA to find the length P

table row cell tan space 32 end cell equals cell fraction numerator P Q over denominator 9 square root of 5 end fraction end cell row cell 9 square root of 5 space tan space 32 end cell equals cell P Q end cell end table 

Work out this value using a calculato

PQ = 12.575... 

Give the answer to the nearest 10 cm (the same as one decimal place)

The height of the tree is 12.6 m

Worked Example

A small wedge consists of three rectangular faces and two triangular faces as shown in the diagram below.

picture-2

Rectangles AEFD and BEFC are congruent.
The base of the wedge, rectangle ABCD lies horizontally.
The maximum vertical height of the wedge is 5 cm and the length of CD is 12 cm.

Find the angle between the planes ABCD and AEFD, giving your answer to one decimal place.

As rectangles AEFD and BEFC are congruent, lengths DF and FC will be equal and so triangle CDF is isosceles
This will also mean the maximum height of the wedge (at point F) lies directly above the midpoint of CD
The angle between the planes required is angle CDF

We can create a 2D right-angled triangle to help solve the problem
Using M as the midpoint of CD and x degree as angle CDF

m3AWf891_picture-1

Use SOHCAHTOA to set up an equation for x degree

tan space x degree equals 5 over 6

Use inverse tan to find x degree

x degree equals tan to the power of negative 1 end exponent open parentheses 5 over 6 close parentheses equals 39.805 space... degree

Round to one decimal place for the final answer

The angle between planes ABCD and AEFD is 39.8° (1 d.p.)

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