3 dimensional problem solving trigonometry

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3D Trigonometry

Here we will learn about 3D trigonometry including how to combine your knowledge of Pythagoras’ Theorem, Trigonometric Ratios, The Sine Rule and The Cosine Rule and apply it to find missing angles and sides of triangles in 3-dimensional shapes.

There are also 3D trigonometry worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is 3D trigonometry?

3 D trigonometry is an application of the trigonometric skills developed for 2 dimensional triangles.

To find missing sides or angles in 3 dimensional shapes, we need to be very clear which rules and formulae we need to use in order to find different angles and side lengths.

The flowchart below can help determine which function you need to use:

You may need to carry out this process several times in a question to fully answer what is being asked. You must be able to justify which rule or formulae you need to use.

Top Tip: Look out for common angles or common sides.

How to use 3D trigonometry to find a missing angle or side length

In order to find a missing angle or side within a 3 -dimensional shape:

Calculate the necessary missing angle or side of a triangle.

• Sketch and label the second triangle using information from Step 1 .
• Calculate the missing angle or side of the final triangle.

Explain how to use 3D Trigonometry to find a missing angle or side length.

3D trigonometry worksheet

Get your free 3D trigonometry worksheet of 20+ questions and answers. Includes reasoning and applied questions.

3D trigonometry examples

Example 1: missing side using trigonometry and pythagoras’ theorem.

The diagram shows a cuboid.

Calculate the length of the line AH . Write your answer in the form a\sqrt{b}\mathrm{cm} where a and b are integers.

The line FH is shared between the two triangles EFH and AFH . We can find the length of FH by using simple trigonometric ratios.

Here we need to find the value of FH (the hypotenuse):

2 Sketch and label the second triangle using information from step 1.

3 Calculate the missing angle or side of the final triangle .

AFH is a right angle triangle, so we can use Pythagoras’ Theorem to find the value of x :

Example 2: finding length using trigonometric ratios

ABCDEF is an isosceles triangular prism. DF is 7cm , Angle DFE is 75 o and angle ECF is 40 o . Find the length of the line CE .

The triangle DEF is isosceles. We can split it in two and find the length of EF by using trigonometric ratios.

We need to find the value of x (the hypotenuse): 

Sketch and label the second triangle using information from step 1.

You must remember not to round your solution too early. Here we will continue to use the full decimal given for x = 13.52296157 .

Calculate the missing angle or side of the final triangle .

Here we have a right angle triangle so we can use another trigonometric ratio to find the length CE :

Example 3:  missing angle including midpoint

ABCDEFGH is a cuboid with the following information:

• The front face of the cuboid is a square.
• The midpoint M lies half-way between E and F .
• The line AC is at 30 o from the line AB .
• BCM is a triangle.
• CE=26\sqrt{3}

Using the diagram below, calculate the size of angle BMC , to 2 decimal places.

The line BC is shared between the two triangles ABC and BCM . We can find the length of BC by using trigonometric ratios.

Here we need to find the value of BC (the opposite side to the angle \theta ):

As M is a midpoint along EF , this gives us an isosceles triangle BCM . As we know CE=26\sqrt(3) this is the vertical height of the triangle.

As we can split an isosceles triangle into two right-angle triangles, and we can use trigonometric ratios to find the angle \frac{\theta}{2} then multiply by 2 to find \theta :

Example 4: missing angle using the cosine rule

ABCDEFGH is a cuboid. BH , FH and BF are straight lines that connect to make a triangle BFH . Using the information in the diagram, calculate the size of angle HFB .

Each side of the triangle BFH is the hypotenuse of one of the three faces of the cuboid. As we have all three dimensions of the cuboid, we can calculate each value ( x, y, z ) as follows:

Here we need to find the value of x (the hypotenuse): 

Here we need to find the value of y (the hypotenuse): 

Here we need to find the value of z (the hypotenuse): 

Keeping BH=\sqrt{106} , we now have the triangle BFH:

As we know the three side lengths of the triangle, we can use the cosine rule to find the missing angle \theta.

Example 5: missing angle using the sine rule

The diagram shows a triangular prism.

• Angle DEC = 50º
• Angle EBC = 55º

The angle ϴ lies between the two lines CE and BE . Calculate the size of angle ϴ . Show all your working.

The line CE is shared between the two triangles CDE and BCE . We can find the length of CE by using trigonometric ratios.

Here we need to find the value of CE (the hypotenuse):

We know two sides and one of the two opposite angles so we need to use the sine rule to find the value for ϴ .

Example 6: missing side using the sine rule

The diagram shows a cylinder:

• A, B and D are points on the circumference of the circles.
• C is the centre of the circle.
• ABC is a triangle
• AD is the diameter of the cylinder.

By calculating the size of angle \theta , work out the height of the cylinder.

As we know two sides and one of the two opposite angles, we can use the sine rule to find the missing angle \theta .

Here we need to find the value of \theta :

\begin{aligned} \sin(\theta)&=\cfrac{\sin(15)}{7.8}\times{25.6} \\\\ \theta&=\sin^{-1}(0.8494573788) \\\\ \theta&=58.15269987… \\\\ \theta&=58.2^{\circ} \; (2dp) \end{aligned}

As AD is the diameter of the circle, we can label AD as 2 × BC = 15.6cm .

As the angle CBD is 90º , we can calculate the angle ABD to equal 90-58.1526…=31.8473…^{\circ}.

We now have enough information on the second triangle to calculate the height of the cylinder. Remember not to round too early.

Calculate the missing angle or side of the final triangle

To find the value of x , we use a simple trigonometric ratio.

Common misconceptions

• Using Pythagoras’ Theorem instead of trigonometry

Using two sides of a non right-angle triangle to find the third side instead of using the cosine rule.

• Incorrect trigonometric ratio used

Incorrect labelling of any triangle can lead to the wrong trig function being used.

• Confusing the Sine Rule with the Cosine Rule

Misunderstanding when to use the sine rule or cosine rule to find a missing side or angle.

• Using the inverse trig function instead and inducing a mathematical error

If the inverse trig function is used instead of the standard trig function, the calculator may return a maths error as the solution does not exist.

Related lessons

3D trigonometry is part of our series of lessons to support revision on trigonometry. You may find it helpful to start with the main trigonometry lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Trigonometry
• Area of a triangle trig
• Cosine rule
• Trigonometric functions

Practice 3D trigonometry questions

1. ABCDEFGH is a cuboid. Calculate the length of DF to 2 decimal places.

Let’s first look at triangle FGH:

We want to know the length of FH which is the opposite. 

\begin{aligned} O&=A \tan(\theta)\\\\ O&=17 \tan(28)\\\\ O&=9.039060338 \mathrm{cm} \end{aligned}

Now that we know the length of FH , we can consider the triangle DFH:

Since we know two sides and we want to calculate the third side, we can use Pythagoras Theorem:

2. ABCDEF is a triangular prism. Calculate the angle DAE . Give your answer to 1 dp.

First we need to look at the triangle DEF:

We need to find the length of DE , which is the opposite.

Now that we know the length of DE, we can consider the triangle ADE :

We want to find the angle DAE. We know O and A .

3. ABCDE is a square based pyramid. By finding the value of x , calculate the perimeter of the base of the pyramid, correct to 2 decimal places.

First we need to work out the value of x:

Now we can look at the base:

The length of the sides can be found using Pythagoras Theorem:

Each side is 70.71m therefore the perimeter is:

4. Three satellites leave Earth on three different trajectories. 2 hours after launch, satellite A is 800km from Earth, satellite B is 500km from Earth and satellite C is 750km from Earth.

Use the cosine rule to calculate the size of angle CAB at this point in time.

AB, AC and BC can all be worked out using Pythagoras Theorem:

We can then apply the cosine rule:

5. Given that GH=10cm, work out the size of the angle GEH . Give your answer to 1 dp.

First we need to calculate the length of EH. We can do this using Pythagoras Theorem:

We can now look at the triangle EGH:

We can calculate angle GEH using the sine rule:

6. ABCDEF is a triangular prism. X, Y, and Z are midpoints on each edge of the prism and triangle XYZ is isosceles. Using this information and the diagram to help you, calculate the size of angle XYZ.

We need to calculate the length XZ using the triangle CXZ:

We can now look at triangle XYZ:

Using the cosine rule:

3D Trigonometry GCSE questions

1.  ABCDEFGH is a cuboid.

Calculate the angle between the diagonal DF and the base AEHD .

Give your answer to 3 sf.

Triangle ADE:

Triangle DEF:

2.  ABCDEF is a triangular prism.

The cross-section of the prism is an isosceles triangle.

M is the midpoint of AC .

Calculate the length of EM .

3.  Find the size of the angle AFH .

Learning checklist

You have now learned how to:

• Apply Pythagoras’ Theorem and trigonometric ratios to find angles and lengths in right-angled triangles (and, where possible, general triangles) in 2 (and 3) dimensional figures

The next lessons are

• Pythagoras theorem
• Alternate angles
• Transformations

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3D Trigonometry

A collection of videos, solutions, activities and worksheets that are suitable for GCSE Maths .

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How to solve problems that involve 3D shapes using trigonometry and the Pythagoras' Theorem?

The following diagram gives an example of the projection of a line on a plane and the angle between a line and and plane. Scroll down the page for more examples and explanations on using Trigonometry and the Pythagoras' Theorem to solve 3D word problems.

GCSE Maths 3 D Trigonometry A* Question

Example: ABCDEFG is a solid cuboid. a) Find the length BE b) Calculate the angle that BH makes with the plane ABFE.

IB Math Studies: 3D Trigonometry

• A room is in the shape of a cuboid. Its floor measures 7.2m by 9.6m and its height is 3.5m. a) Calculate the length of AC b) Calculate the length of AG c) Calculate the angle that AG makes with the floor.
• The right pyramid shown in the diagram has a square base with sides of length 40 cm. The height of the pyramid is also 40 cm. a) Find the length of OB. b) Find the size of angle OBP.

Trigonometry and Pythagoras in 3D Shapes Calculating an angle between an edge and plane in a 3D shape using Pythagoras and trigonometry.

Example: For this cuboid calculate a) Length AG b) Angle between AG and the plane ABCD

Angle between a line and a plane AddMaths Ex8A Q1

3D Trigonometry Problem The worked solution to a three-dimensional trigonometry problem. A tree on the far side of a river bank is used to determine the width of the river with the help of a few right angle triangles. Step 1: Identify the right angles in your diagram. Step 2: Include all the measurements in your diagram.

How to do Trigonometry in Three dimensions? 3D trig Pythagoras cuboid Part 1

Example: ABCDEFGH is a cuboid with dimensions 4m, 2m and 1.5m. X is the midpoint of the side EF. Find the lengths AC, AG and AX. Find the angles GAC and AXB.

3D Trig Pythagoras cuboid part 2.

3D Trig Pythagoras cuboid part 3.

3D Trig Pythagoras cuboid part 4.

3D Trig Pythagoras cuboid part 5.

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This is level 1: Three dimensional problems which can be solved using Pythagoras' theorem. You can earn a trophy if you get at least 4 questions correct and you do this activity online . Give answers correct to three significant figures.

This is Trigonometry in Three Dimensions level 1. You can also try: Level 2 Level 3 Level 4 Level 5

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Pythagoras' Theorem - Start by applying Pythagoras' theorem in two dimensions.

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Level 1 - Three dimensional problems which can be solved using Pythagoras' theorem

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Lesson: Trigonometry in Three Dimensions Mathematics

In this lesson, we will learn how to solve trigonometry problems in three dimensions.

Lesson Plan

Students will be able to

• recognize where to apply the trigonometric ratios sine, cosine, and tangent to solve problems involving angles in three dimensions,
• find planar right triangles in three-dimensional figures to solve trigonometric problems,
• combine trigonometry and the Pythagorean theorem to solve problems involving angles and lengths in three dimensions,
• solve three-dimensional trigonometry problems in real-world contexts.

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Solving 3D Problems using Trigonometry

April 4, 2021.

When teaching solving 3D problems using trigonometry we begin the lesson with a recap of Pythagoras ’ Theorem and the three trigonometric ratios.  We do this by matching the ratio and equations to the respective right-angled triangle.  

Students are encouraged to work in pairs and to show the diagrams as part of the working out on mini-whiteboards.   A nice way to differentiate this is to have the less able student do the writing.  I check to see whether they can correctly label the sides as this helps t o know whether to use Sine, Cosine or Tangent.  

Solving 3D Problems using Trigonometry 

When solving 3D problems involving trigonometry, I demonstrate how to identify the right-angled triangle containing the unknown side or angle and sketch it out separately.  If there is not enough information given about this triangle, we look for further clues in another triangle involving the same line or angle.  Check out the video below for a quick demonstration.  

Identifying the Right-Angled Triangles 

To check the student’s understanding and progress we work through some examples from the Interactive Excel File on mini-whiteboards.    

Students are asked to sketch out the necessary right-angled triangles needed to find the length FB and angle FBG.  Without explaining how, I ask them to calculate the length FB and show me their working on mini- whiteboards.  It is interesting that a couple of students try to apply the Cosine Rule to triangle FBG.   We feedback how to use Pythagoras’ Theorem to address the misconceptions.  

Next, I ask students to use this new information to calculate angle FBG.  Some struggle to use the correct ratio but all students attempt to solve it as a right-angled triangle using either Sine, Cosine or Tangent.  

I use the solving 3D problems using trigonometry Interactive Excel File to pose a different, but similar question for the class to attempt again on mini-whiteboards.  All students sketch the two relevant right-angled triangles.  S ome continue to  struggle calculating the angle using the correct trigonometric ratio but all students correctly calculate the length FB using Pythagroas ’ Theorem.    

Applying Trigonometry to Square Based Pyramids 

The plenary challenges the class to identify the necessary right-angled triangles independently.   I provide a handout of the shape to some students so they can annotate the drawing.  After a couple of minutes of attempting the problem independently we stop the lesson to discuss different approaches.  This provides a starting point for some students while reassuring others.  The plenary problem normally takes between 10 to 12 minutes.  To end the lesson , I ask a willing student to demonstrate their working at the front of the class to their peers.  We address any misconceptions as they arise.

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About mr mathematics.

My name is Jonathan Robinson and I passionate about teaching mathematics. I am currently Head of Maths in the South East of England and have been teaching for over 15 years. I am proud to have helped teachers all over the world to continue to engage and inspire their students with my lessons.

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Three-dimensional trigonometry problems

Three-dimensional trigonometry problems can be very hard and complex, mainly because it’s sometimes hard to visualise what the question is asking.  If there is a diagram given in the question it can make things easier, but it can still be challenging thinking about exactly what you need to do to find an answer.  It pays to keep a calm head and think about exactly what you need to find and all the things that you know . 

Usually these 3D questions will require more than one step to solve.  For instance, take this classic pyramid type question, where you need to find the length AE:

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3D diagrams are usually more complex than 2D ones and have more information present in the diagram.  This one shows a pyramid.  The four dashes through the four sides of the base show that all four sides are equal in length, and the right angle shown in one corner of the base tells us for certain that the base is a square (if all four sides were equal length but there were no angles marked it could be a parallelogram base).

The vertical height of the pyramid (not the slant height) is 10 units.  Also we know that the angle formed between the slant line up the middle of a side and the base is 72°.

Here’s where we ask ourselves – what are we trying to find?  In this case, it’s given straight to us in the question – the length of side AE.  How can we find out this length?  Well, there are a couple of triangles which it forms part of, shown here:

Both these triangles are right angled triangles , meaning if we could find the length of both other sides, we could find the length of side AE using Pythagoras’ Theorem.  The second triangle option looks attractive, because one of the sides is the vertical part of the triangle, which we already know is 10 units long.  So let’s work with this triangle.  We just need to find the length of its short side now.

The short side of the triangle forms half of the diagonal of the square base.  At this stage it pays to add some more letters into our diagram so we can more easily describe what we’re doing in any calculations.  We can also draw a square representing the base, and a triangle representing the one we’re working with in the diagram:

So to recap – we know that EF is 10 units long.  If we can work out the length of AF, then we can use Pythagoras’ Theorem to work out the length of AE.  So how do we find out AF?  Well, look at the square base diagram.  If we can work out the side length of the square base, we can use Pythagoras’ Theorem to work out the length of AF.  So let’s work on the side length of the base.

The only unused piece of information in the question is the 72° angle.  Let’s have a look at that triangle:

Its short side forms part of the square base.  In fact, its short side is exactly half the length of the base’s sides.  We can use trigonometry to work out the length of FG, which is half the side length of the base.  Here’s the triangle we’re working with:

If we go back to our diagram of the base, we now know the length of FG:

The two lengths marked with arrows are the same length as FG, so we can use Pythagoras’ Theorem to work out the length AF:

We can go back to our original triangle with the side AE in it that we’re trying to find.  We now know the lengths of the other two sides (AF and EF), so it’s just Pythagoras’ Theorem to find AE:

Notice how many 2-dimensional diagrams I drew along the way.  It makes it easier when you start off with the 3-dimensional diagram, and then work at bits of it, by drawing 2-dimensional parts of it and working on each part individually.  By labelling the corners and important parts of the 3-dimensional diagram, you can also label the points in these simpler 2-dimensional diagrams.  Having labels in both diagrams will help you keep track of where the simple 2D diagrams fit into the 3D one.

Copyright 2007-2011 Michael Milford