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3D trigonometry Simplified Revision Notes

Revision notes with simplified explanations to understand 3D trigonometry quickly and effectively.

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3D trigonometry

Introduction to 3D Trigonometry

3D Trigonometry is quite similar to regular 2D trigonometry, but with an added dimension! The key to solving 3D problems is to visualize and break them down into simpler 2D problems involving right-angled triangles. Here's how you can tackle 3D trigonometry problems step by step.

The Secret to Solving 3D3D Trigonometry Problems

  1. Understand that 3D Trigonometry uses the same principles as 2D2D Trigonometry:
  • Pythagoras' Theorem: To calculate the length of a side in a right-angled triangle.
  • Sine, Cosine, and Tangent (Sin, Cos, Tan): To calculate angles and lengths in right-angled triangles.
  1. The Challenge in 3D3D Trigonometry:
  • The hardest part is often just spotting where the right-angled triangles are within the 3D3D shape.
  1. Steps to Solve a 3D3D Trigonometry Problem:
  • Draw it out flat: Reduce the 3D3D problem to a 2D2D problem by projecting the relevant triangle onto a flat plane.
  • Label your sides: Identify and label the sides of the triangle as Hypotenuse (HH), Opposite (OO), and Adjacent (AA).
  • Fill in the information you know: Insert the given lengths and angles.
  • Work out what you don't know in the usual way: Apply Pythagoras' Theorem or the trigonometric ratios (Sin, Cos, Tan) to find the missing side or angle.

Worked Example: Finding Distances in a 3D3D Shape

infoNote

Example

Problem: The diagram shows a wedge of Cheddar cheese, where rectangle PQRS is perpendicular (at 90°) to rectangle RSTU. The distances are PQ = 2.5m, QR = 7.8 m, and TU = 4.9 m. You need to calculate:

  • (a) The distance QTQT
  • (b) The angle QTRQTR

Step-by-Step Solution:

  1. Understanding the Problem:
  • You need to find the distance QTQT, which is a line running through the 3D3D shape, and the angle QTRQTR.
  • To do this, you'll need to identify right-angled triangles within the shape.
  1. Finding QTQT:
  • Step 1: Spot the right-angled triangle involving QTQT. In this case, it's triangle QTR.
  • Step 2: Break down the problem into two triangles:
  • First, calculate TRTR using the triangle TRUTRU.
  • Then use the triangle QTRQTR to find QTQT.
  1. Calculating TRTR:
  • Given: TRTR is the hypotenuse of triangle TRUTRU.

  • Use Pythagoras' Theorem:

TR2=TU2+UR2TR^2=TU^2+UR^2TR2=4.92+7.82=24.01+60.84=84.85TR^2=4.9^2+7.8^2=24.01+60.84=84.85TR=84.85=:success[9.21m]TR=\sqrt{84.85}=:success[9.21 m]
  1. Calculating QTQT:
  • QTQT is the hypotenuse of the triangle QTRQTR.
  • Use the lengths QRQR and TRTR:
QT2=QR2+TR2QT^2 = QR^2 + TR^2 QT2=2.52+9.212=6.25+84.85=91.1QT^2 = 2.5^2 + 9.21^2 = 6.25 + 84.85 = 91.1 QT=91.1:success[9.54m]QT = \sqrt{91.1} \approx :success[9.54 m]
  1. Finding Angle QTRQTR:
  • Using Tan:
tan(θ)=QRTR\tan(\theta) = \frac{QR}{TR} θ=tan1(2.59.21):success[15.2°]\theta = \tan^{-1} \left( \frac{2.5}{9.21} \right) \approx :success[15.2°]

Final Answers:

  • Distance QTQT: Approximately 9.54 m.
  • Angle QTRQTR: Approximately 15.2°.

Worked Example: Calculating the Height of a Vertical Pole in a Tent

infoNote

Example Problem: A tent is shown in the diagram with a vertical pole OP. The pole is at the centre of rectangle QRSTQRST. Given that TQ=13 m, TS=5 m, and SR=12 m, find the height of the pole OPOP.

Step-by-Step Solution:

  1. Breaking Down the Problem:
  • Objective: Calculate the height OPOP of the vertical pole.
  • Identify relevant triangles and determine the lengths or angles that can be used to find OPOP.
  1. Finding the Length OTOT:
  • OTOT is half the length of TRTR, since OO is the midpoint of TRTR.
  • Step 1: Calculate TRTR using Pythagoras' Theorem in the right-angled triangle TRSTRS

Calculation:

TR2=TS2+SR2=52+122=25+144=169TR^2=TS^2+SR^2=5^2+12^2=25+144=169TR=169=:success[13m]TR=\sqrt{169}=:success[13 m]
  • Therefore, OT=TR2=132=:highlight[6.5m]OT=\frac{TR}{2}=\frac{13}2=:highlight[6.5m].
  1. Calculating the Height OPOP
  • Now, consider the right-angled triangle OTPOTP, where OT=6.5 m and ∠OTP=48°.
  • Use the tangent ratio to find OPOP. Calculation:
tan(θ)=OppositeAdjacent=OPOT\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{OP}{OT} tan(48)=OP6.5\tan(48^\circ) = \frac{OP}{6.5} OP=6.5×tan(48)=6.5×1.1106:success[7.22m(2dp)]OP = 6.5 \times \tan(48^\circ) = 6.5 \times 1.1106 \approx :success[7.22 m (2dp)]
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