3D Trigonometry and Pythagoras - GCSE Higher Maths

Summary

In this video, the creator, 1st Class Maths, dives into the concept of 3D trigonometry and Pythagorean theorem, building on foundational knowledge of these topics in 2D. The video illustrates the methodology of finding right-angled triangles within 3D shapes like cuboids and triangular prisms, emphasizing the importance of labeling sides and using the appropriate formulas. Examples are provided to demonstrate practical applications, such as calculating lengths and angles, offering a step-by-step guide to solving complex GCSE Higher Maths problems. The video lacks a direct outline for beginners and assumes a pre-acquired understanding of 2D principles, dedicating the content to more complex problem-solving scenarios.

Highlights

• Learning 3D trigonometry is a subscriber-requested topic, indicating high interest! 😀
• The video reinforces the necessity of understanding 2D shapes to grasp 3D concepts. 🧠
• Practical examples demonstrate finding right-angled triangles in 3D shapes. 📏
• Emphasis on using Pythagoras' theorem for calculating hypotenuse in 3D. 📐
• The creator cautions against premature rounding to ensure accuracy. ✅

Key Takeaways

• Mastering 2D trigonometry and Pythagoras is essential before tackling 3D problems! 🎓
• Cuboids and triangular prisms are common shapes in 3D trigonometry problems. 🧊
• Identifying the right-angled triangle in the 3D shape is crucial for solving these problems. 🔍
• Each problem may involve different strategies, including using Pythagoras' theorem and trigonometry. ➕
• Avoid premature rounding for accuracy in calculations; always use exact values until the final answer. 🚀

Overview

This video, as requested by a subscriber, delves into 3D trigonometry and Pythagoras, aiming to enhance your problem-solving toolkit. It begins by refreshing your memory on the fundamental principles of 2D trigonometry, setting a solid foundation for more advanced calculations in three dimensions.

We explore various 3D shapes, particularly focusing on cuboids and triangular prisms, to understand how right-angled triangles appear within these shapes. The approach combines Pythagoras' theorem and trigonometric functions to navigate through complex geometrical problems, fostering a deeper understanding and application of mathematical techniques.

With insightful examples and a step-by-step breakdown, this video is a practical guide for GCSE Higher Maths students striving to perfect their grasp of trigonometry and geometric calculations. The tutorial urges learners to focus on precision, especially regarding rounding practices, ensuring they retain accuracy until the very last step.

Chapters

• 00:00 - 00:30: Introduction to 3D Trigonometry and Pythagoras The chapter introduces 3D trigonometry and Pythagoras, starting with a basic understanding of 2D concepts. It emphasizes the necessity of being proficient in 2D Pythagoras and trigonometry before advancing to 3D. The introduction is set against a backdrop of interactive engagement with viewers, encouraging them to suggest topics. The foundational approach begins with the basics of right-angled triangles, labeling the sides and understanding the relationships between them as a preparation for delving into 3D complexities.
• 00:30 - 01:00: Basic 2D Concepts In the chapter titled 'Basic 2D Concepts', the transcript discusses foundational concepts of trigonometry involving right-angle triangles. It describes the relationship between the sides of a triangle using the Pythagorean theorem, A² + B² = C², where C is the hypotenuse. The chapter further explains how to label the sides of a triangle with respect to an angle X: 'hypotenuse' (opposite the right angle), 'opposite' (facing the angle X), and 'adjacent' (next to the angle X). Additionally, it outlines basic trigonometric functions: Sine (opposite/hypotenuse), Cosine (adjacent/hypotenuse), and Tangent (opposite/adjacent).
• 01:00 - 02:00: Finding Right Angle Triangles in a Cuboid The chapter discusses strategies for identifying right angle triangles within a cuboid. It highlights the importance of being familiar with cuboid properties, suggesting a revision of foundational topics if necessary, and promising useful resources in the video's description. The chapter introduces the concept of labeling the cuboid's vertices as a standard method used across various problems. The narrative points out that most questions will involve finding 2D right angle triangles, which can be discovered on the cuboid's faces. An example is given where vertices a, e, and d are connected to form such a triangle.
• 02:00 - 03:00: Example: Finding Length of DG The chapter discusses the concept of visualizing and identifying right-angle triangles within different faces of a 3D geometric shape. It starts by explaining how a right-angle triangle can be drawn on the front face by connecting vertices A, B, and D. The narrative expands on how these triangles can be outlined on various faces or even imagined internally through the shape, using different vertex connections such as B to F to D. The emphasis is on creatively recognizing these triangles in various dimensions and planes within geometric figures.
• 03:00 - 04:00: Example: Finding Length of AC The chapter discusses different ways to identify and solve for the length of a segment in a geometrical shape, specifically focusing on recognizing right angle triangles that may not be immediately apparent. It highlights strategies for tackling questions that involve triangles passing through shapes, making them harder to identify due to the positioning of right angles.
• 04:00 - 05:00: Example: Finding an Angle Using Trigonometry The chapter focuses on solving a geometric problem using trigonometry to find a missing length in a triangle. The task involves identifying the length of line DG using the triangle DCG as the context, where DG represents the hypotenuse. The method involves illustrating the triangle, identifying known side lengths, and applying the Pythagorean theorem to compute for DG.
• 05:00 - 06:00: Harder Example: Finding Length of AG The chapter focuses on calculating the length of line AG. The method involves using the Pythagorean theorem, demonstrated with a different line segment DG. The segment DG is solved by finding the square root of the sum of squares (3^2 + 4^2 = 25), resulting in DG equating to 5 cm. The chapter continues by addressing how to determine the length of another line, AC, emphasizing the visualization of right-angled triangles within a geometric figure (a cuboid).
• 06:00 - 08:00: Working with Triangular Prisms In this chapter, we explore working with triangular prisms, specifically focusing on calculations involving the sides of triangles. The discussion centers around using Pythagorean theorem to find the hypotenuse of triangle ADC. Given the base=10 and height=3 of the triangle, we apply the formula a² + b² = c² to find AC (hypotenuse). The resulting equation is AC² = 10² + 3² = 109. To obtain the length of AC, take the square root of 109, which requires a calculator and should be expressed to one decimal place. This method highlights the geometric and arithmetic techniques essential in dealing with triangular prisms.
• 08:00 - 10:00: Example: Finding Volume of a Square Base Pyramid The chapter focuses on calculating the volume of a square base pyramid. It involves using trigonometry to find sides and angles in cuboids which may be components of the pyramid. Specifically, the chapter details how to find the length of segment AC using a calculator, rounding it to one decimal place (10.4 cm). It then goes on to discuss finding an angle FCG by analyzing a triangle formed by the points F, C, and G, showing its relation in the structure of the cuboid.

3D Trigonometry and Pythagoras - GCSE Higher Maths Transcription

• 00:00 - 00:30 [Music] this video's topic has been requested by one of my subscribers if there's a topic you'd like to see in the next video just subscribe and leave me a comment so in this one we're going to look at 3D trigonometry and Pythagoras in order to do this topic you need to be good at both of these in two Dimensions first for 2D Pythagoras we take a right angle triangle we label the two shorter sides as a B and the longest side the
• 00:30 - 01:00 hypotenuse is C we can then use the formula A2 + B2 = c^2 to find one of the missing sides for 2D trigonometry we take a right angle triangle and if we label the angle X we then label the sides the hypotenuse opposite the right angle the opposite opposite the angle given and the adjacent is next to the angle given we can then write down that the sign of the angle is the opposite divided by the hypotenuse the COS of the angle is the adjacent divid by the hypotenuse and tan of the angle is equal to the opposite divid by the adjacent if
• 01:00 - 01:30 you aren't comfortable with either of these topics I'd suggest you go and revise those first I'll put some links in the video's description for this topic the most common shape in exam questions is a cuboid we can label the vertices of this cuboid a b c d e f g h in all of the questions that we're going to look at in this video we're going to need to find two dimensional right angle triangles and there are loads that you can find in this shape firstly you could find a right angle triangle that's on any of the faces so if I connect up a to e and then e to D and then back to a we have a
• 01:30 - 02:00 right angle triangle on this front face here we could draw a right angle triangle on any of the faces so if I connected A to B and then across to D and then back to a there's now a right angle triangle on the bottom face we could even do one on the face on the side here from C to g g to D and back to C and there's a right angle triangle here some of the right angle triangles won't be on the outside faces of the shape sometimes they'll go through the shape so for instance if I connected B to F and then F to D and then back back
• 02:00 - 02:30 to B again this would give me a right angle triangle but this one goes through the shape you can see the right angle is down at B you could also have one that starts at B goes across to G and then over to H and back to B again this one's much harder to see but it is a right angle triangle the right angle is up here at G when the triangle goes through the shape rather than being on one of the outside faces it's usually a more difficult question now let's have a look at some of the questions you may be asked so if we label on some of the lengths here we're going to start with some easier
• 02:30 - 03:00 questions work out the length of DG to work out the length of DG the first thing to do is draw a line connecting D and G next we're going to look for a right angle triangle that involves that line since the line DG is just on this outside face here it makes sense to choose this triangle the triangle dcg so the triangle dcg looks like this and we have the length on the bottom from D to C which is three and the height from C to G is four we're trying to find the length of DG here which is the hypotenuse so we can just use pythag
• 03:00 - 03:30 so we could say that dg^ 2 is 3^ 2 + 4^ 2 if you do 3^ 2 + 4 S you end up with 25 so dg^ squ is 25 and then to get DG you square root both sides so DG would equal 5 cm so that's that question done DG is 5 cm what about if we were asked to work out the length of AC to one decimal place once again we start by just drawing the line from a to c now we look to see if we can find a right angled triangle that involves that line I can see one on the Bas of the cuboid here
• 03:30 - 04:00 the triangle ADC so if I draw out triangle ADC it looks like this and we know the base is 10 and the height is three we're after the line AC which is also the hypotenuse so once again we use Pythagoras so ac^ 2 is 10 2 + 3^ S if you do 10 s + 3 squ you get 109 so a squ is 109 and to find AC you square root both sides so AC will be the square root of 109 you'll need a calculator for this one and you can tell because it says to give it to one decimal place if you type
• 04:00 - 04:30 this into your calculator you'll find AC is this number here and to round it to one decimal place that would be 10.4 CM sometimes rather than working out sides we need to work out angles so let's say we were asked to find the angle F CG to find the angle fcg we need to look at the triangle that involves F C and G you can see we've already got FG and GC marked on they're the sides of the cuboid so we need to connect up from FC as well and it's this triangle here so if we draw out the triangle fcg it looks like this and looking at our cuboid we
• 04:30 - 05:00 can see from C to G is 4 cm and we actually know from F to G as well since that's the same as from a to d so that's 10 cm we need to try and find the angle fcg that's the one in the bottom right so I'm going to label that as X now this time we're going to use trigonometry rather than Pythagoras since it has an angle to do this we label up the sides you can see the one opposite the angle is the 10 cm so that's the opposite the one that's next to the angle is the 4 cm that's the adjacent since we've got the opposite and the adjacent we're going to use tan so we can write down that tan of
• 05:00 - 05:30 the angle so tan of X is equal to the opposite which is 10/ by the adjacent which is 4 in this one we're trying to find the angle so we're going to use inverse tan 10 ID 4 is just 2.5 so we could say that X is equal to inverse tan of 2.5 type this into your calculator and you'll get this number here this question has asked us to round it to one decimal place again so 68.2 de so in all of these questions so far the key thing to do is to try and find the right angle triangle that involves the information we've been asked for in the question
• 05:30 - 06:00 sometimes the questions are a little bit more tricky though let's have a look at a harder one so if we add some lengths to this cuboid here and we're going to try and find the length of AG to one decimal place AG is this line here notice this line isn't on one of the faces of the cuboid it goes through the shape so to find the right angle triangle for this one we're going to have a triangle that goes through the shape again to show this triangle I'm going to connect from a to c as well and you can see that the triangle here goes through the shape with the right angle in the bottom right corner at C so if we
• 06:00 - 06:30 draw out that triangle or look like this triangle ACG and we know the length from C to G that's 20 cm but that's actually all the information we know so if we're going to try and find AG using this triangle we're going to need to find some more information first for instance the length from a to c so how could we find the length from a to c well to do this we need to look at a different right angle triangle first the right angle triangle we need is actually on the base of the cuboid this one here a d c so if we draw our triangle ADC it
• 06:30 - 07:00 looks like this and we know two of the lengths for this triangle D to C is 8 cm and a to d is 9 we can use this information to find AC which will allow us to find AG G in the red triangle so if we use Pythagoras to find the length of AC we would say that AC s is 8 s + 9 s if you do 8 squ + 9 squ you get 145 and then if you square root both Sid you'll find the value of AC is the < TK of 145 which is this number here so we've now found AC we can return to the
• 07:00 - 07:30 original red triangle this one here we know that AC is 12.04 159 and so on now since we're going to use that number in a moment we're not going to round it off that's called premature rounding where you round the number before you finish the question a much safer thing to do here is rather than writing AC as 12.04 and so on is we write it as < TK 145 the reason this is safer is because < TK 145 is the exact length of AC now we can go ahead and find AG by
• 07:30 - 08:00 using Pythagoras once again on this triangle to do this we would say that AG ^2 is equal to 202 plus squ < TK of 145 all squared and if you type all of that into your calculator you'll end it with ag ^2 = 545 now to find AG just square root both sides so AG will be the square < TK of 545 and if you type this into your calculator you get this number here the question wants the answer to one decimal place and now that this is the end of the question we can round the answer so it's 20 3.3
• 08:00 - 08:30 CM sometimes the shape you're given in the question isn't a cuboid it could be a triangular prism like this one here for this question we're going to work out the size of the angle between EC and the plane ABCD so first of all let's draw on the line EC which is this line here and what does it mean by the plane AB c d well a plane is simply a flat surface so we need to find the flat surface that connects a b c and d that's just the base of the triangular prism here but it's still not immediately clear from our diagram which angle we're
• 08:30 - 09:00 trying to find if you're ever trying to find the angle between a line and a plane you'll notice that one of the points will be on the plane and one of them isn't you can see that c is on the plane but e is above the plane what you want to do is drop a vertical line down from E until you hit the plane so that would be down here and then connect up the two points that are on the plane so A and C across here you can now see that the angle between the line EC and the plane is this angle in here so let's label that One X the type of wording used in this question is common for Ed
• 09:00 - 09:30 Exel but not AQA if it were an AQA question it would probably just say work out the size of angle AC so now let's go and solve this question if I look at the triangle AC I don't actually have any of the information for this triangle at all so I'm going to need to work out at least two bits before I can return to it to find X first of all I'm going to look at the triangle that's on the front of the prism here this will allow me to find the length from a to e so let's look at the triangle a d e we know the angle is 20° we also know the length from E to D
• 09:30 - 10:00 because that's the same as the length from F to C so that must be 14 cm we can now use trigonometry to find a to e so let's label on AE will be the opposite and E is the hypotenuse and we don't need a d That's the adjacent so for this one we use the opposite and the hypotenuse so we need to use S so we could say s of the angle so s of 20 is equal to the opposite which we don't know so we'll just label that as EA and divide this by the hypotenuse which we do know that's 14
• 10:00 - 10:30 if we multiply both sides by 14 we find that EA is equal to 14 sin 20 which if you type into your calculator gives you this number here so now we found the length from e to a we can add this information to the red triangle so EA is 4.78 and so on now in order to find X we still need more information from this triangle we're going to try and find the length from E to C next to do this we're going to use the triangle from E to F to C this is on the top face of the prism so if we draw out this triangle it looks like this we know the length from F to C
• 10:30 - 11:00 is 14 and we actually know the length from E to F that's the same as from D to C so that's 10 we can use Pythagoras to find the length from E to C so if we do ec^ 2 is 10 2 + 14 s if you use your calculator for this you'll find that EC squ is equal to 296 and then to get EC we just square root both sides so EC will be the square root of 296 now you could square root this and turn it into a decimal but since I know I'm going to need this length in a moment I'm I'm going to leave it in that third form as < TK 296 so if I go back
• 11:00 - 11:30 over to my red triangle now I can label from E to C as < TK 296 now we're ready to use this triangle to go ahead and find X so for this one it's trigonometry we label the opposite over here EA and we can label the hypotenuse that's < TK 296 and for this one we need to use S so s of X is the opposite so this 4.78 and so on number divide by the < TK of 296 now you need to be really careful when you type that right hand side into
• 11:30 - 12:00 the calculator you cannot just type 4.78 that would be another example of premature rounding since this isn't the answer to the question yet we can't round so you need to make sure that you get that 4.78 back up on your calculator with all of the decimals after it so we got the 4.78 answer on by doing 14 sin 20 so do 14 sin2 and that number will be on your screen divide that by the sare < TK of 296 and you get this number here if you did just use 4.78 then you may get a slightly different answer to the
• 12:00 - 12:30 real answer so now we can use inverse sign to find the answer to this angle we do xals inverse sign of this number here which gives us the answer 16.2 when rounded to one decimal place another shape that you sometimes see in exam questions is a square Bas pyramid we may get a question that says ABCDE e is a square base pyramid and we might be asked to work out the volume of the pyramid to find the volume of a pyramid you use the formula 1/3 * the base area time the perpendicular height now at the moment we don't know the
• 12:30 - 13:00 perpendicular height and we also don't know the base area so we're going to need to go and find those first and then return to this formula later let's start by trying to find the perpendicular height which would be from M to e to do that we need to use this triangle here so if we take a closer look at this triangle we want to find the length from M to e this will be using trigonometry so if we label up this triangle the one opposite the 72 that's the opposite one opposite the right angle is the hypotenuse and the other side is the adjacent the the perpendicular height is from M to e
• 13:00 - 13:30 that's the opposite and we have the hypotenuse so we're going to use S so s of 72 will equal the opposite which we don't know so we can just call that me over the hypotenuse which is 15 to get me then you multiply both sides by 15 so me is 15 sin 72 which is this number here now it will also be helpful in this question to work out the length from M to C since MC is the adjacent and we know the hypotenuse we can use C so COS of 72 is equal to the adjacent so MC divide by the hypotenuse which is 15 if
• 13:30 - 14:00 you multiply both sides by 15 you find that MC is equal to 15 cos 72 which is this number here so if we add both of these lengths onto our diagram we've got the height now of me which is 14.2 and so on and we've also got MC which is 4.6 and so on so we've now got the perpendicular height but we need the base area we're going to look at this triangle here that's on the base of the pyramid so if we draw out this triangle ADC so we don't know any of the length of this triangle at the moment but we do
• 14:00 - 14:30 know there's a point M right in the middle of the line AC and to get from M to C is 4.6 and so on now since m is the midpoint of the line a the length from a to M must be the same as the length from M to C so from a to m is also 4.6 and so on so if we wanted to find the length from a all the way to C we would just double this number now remember we're not doubling 4.6 here we're doubling that whole number 4.6 and so on with all of the remaining digits so if you get that number back back on your calculator screen and double it you'll end up with
• 14:30 - 15:00 9.27 and so on now this is where things get a little bit tricky we want to find the length of a and DC and it doesn't look like we have enough information to find them since we only have one of the sides but it is possible this pyramid is a square base pyramid this means the length from a to d must be the same as the length from D to C since the base of the pyramid is a square so whatever ad is let's call it X that's the same as DC so that's also X so this means the hypotenuse squared so
• 15:00 - 15:30 9.27 and so on squared will be the same as the sum of the squares of the other two sides so X2 add X2 if we leave the left hand side alone but then add X2 and X2 we get 2 x^2 if you do square the 9.27 and so on and then divide it by two you'll end up with this number here and if you divide the right hand side by two you get x s to get X we can then just square root both sides so X will be the square root of this number which gives you this number here so
• 15:30 - 16:00 we've now found the lengths of the square on the base so from a to d is 6.5 and so on and so is D to C we're now ready to finally find the volume of the pyramid let's go back to that formula the volume is 1/3 * the base area time the perpendicular height so we need to do 1/3 multiplied by the area of the base to find the area of the square you just multiply its length and its height so 6.5 and so on time 6.5 and so on and then we multiply this by the perpendicular Heights which we found earlier that's from M to e 14.2 and so
• 16:00 - 16:30 on once again I cannot stress how important it is that you don't use rounded numbers here you cannot type 6.5 * 6.5 * 14.2 you need to use the four number that was originally on your calculator screen so if you do that properly using that whole number you'll end up with this number here and since this is the answer to the question we are able to round this one hasn't actually given us an accuracy so let's say it was to one decimal place 24.3 cm Cub thank you for watching this video I
• 16:30 - 17:00 hope you found it useful check out the one I think you should watch next subscribe so you don't miss out on future videos and go and try the exam questions in this video's description