The shape of the lesson
By the end of the lesson, learners can
- Lengths in a cuboid: In the lab, rotate the cuboid and switch on the base diagonal, then the space diagonal.
- The angle between a line and a plane: Shine a light straight down onto the plane: the line's shadow is its projection.
- Problems in context: Rooms, fields with flagpoles and tents are all cuboids, rectangles and pyramids in disguise.
- Label the solid and mark every right angle (vertical meets horizontal at \(90^\circ\)).
- Base first: find the diagonal or distance in the base with Pythagoras.
- Then upright: use the vertical triangle for the slant length or the angle.
- Line and plane: drop a perpendicular; the angle is between the line and its projection.
- Accuracy: keep surds or full calculator values until the end.
One picture every idea returns to
Work in flat right-angled triangles. The space diagonal of a cuboid is \(\sqrt{l^2 + w^2 + h^2}\). The angle between a line and a plane is the angle between the line and its projection (its 'shadow') on the plane, found in the right-angled triangle formed by dropping a perpendicular to the plane.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Find the flat triangle inside the solid.. Extended. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Lengths in a cuboid | In the lab, rotate the cuboid and switch on the base diagonal, then the space diagonal. Project tab 6 of the lab (Sine and cosine rules lab, 3D problems) and let a student drive it while the class predicts. Model the worked example The space diagonal on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The angle between a line and a plane | Shine a light straight down onto the plane: the line's shadow is its projection. Model the worked example Diagonal and base on the board, one line at a time. Your turn: A cube has edges of 4 cm. Find (a) the exact length of a space diagonal (b) the angle between a space diagonal and the base. | Pairs |
| 21 to 30 min | Teach and check: Problems in context | Rooms, fields with flagpoles and tents are all cuboids, rectangles and pyramids in disguise. Model the worked example Flagpole in a field on the board, one line at a time. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 19 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Problems in context, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Using half the side instead of half the DIAGONAL for a pyramid's sloping edge. \(M\) is the centre of the square, so \(AM\) is half of \(AC\), not half of \(AB\). | (Half the side is correct only for the slant height up the middle of a face.) |
| Using the wrong line in the base. | The angle between \(AG\) and the base is NOT angle \(GAB\): \(AB\) is not the shadow of \(AG\). Using \(\tan^{-1}(5/8) = 32.0^\circ\) is the classic wrong answer. |
| Measuring the angle of elevation from a ground line that does not lead to the foot of the pole. | The angle of elevation of \(P\) from \(C\) lives in the vertical triangle \(PAC\), so you need \(AC\), the line from \(C\) to the foot \(A\). |
Pythagoras and trigonometry in 3D: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
A cuboid measures 7 cm by 5 cm by 4 cm. Calculate the length of a space diagonal.
For the cuboid in the previous question, with base 7 cm by 5 cm and height 4 cm, calculate the angle between a space diagonal and the base.
Without a calculator, show that the space diagonal of a cube of edge 6 cm is \(6\sqrt{3}\) cm.
A pyramid has a rectangular base \(ABCD\) with \(AB = 10\) cm and \(BC = 6\) cm. Its vertex \(V\) is 8 cm vertically above the centre \(M\) of the base.
(a) Calculate the length of the edge \(VA\).
(b) Calculate the angle between \(VA\) and the base.
A rectangular field \(ABCD\) has \(AB = 40\) m and \(BC = 30\) m. A vertical mast \(AP\) of height 15 m stands at \(A\).
Calculate the angle of elevation of \(P\) from \(C\).
A room is a cuboid 5 m long, 4 m wide and 3 m high. A string is stretched from a corner of the floor to the opposite corner of the ceiling.
(a) Calculate the length of the string.
(b) Calculate the angle between the string and the floor.
(c) Show that the angle between the string and one of the 4 m by 3 m end walls is exactly \(45^\circ\).
Answer key and marking notes
- \(d = \sqrt{7^2 + 5^2 + 4^2} = \sqrt{49 + 25 + 16} = \sqrt{90}\)
- \(d = 9.49\) cm (3 s.f.)
- Base diagonal \(= \sqrt{7^2 + 5^2} = \sqrt{74}\) cm
- \(\tan\theta = \dfrac{4}{\sqrt{74}}\)
- \(\theta = 24.9^\circ\) (1 d.p.)
- \(d^2 = 6^2 + 6^2 + 6^2 = 108\)
- \(d = \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\) cm
- \(AC = \sqrt{10^2 + 6^2} = \sqrt{136}\), so \(AM = \tfrac{1}{2}\sqrt{136} = \sqrt{34} = 5.830\ldots\) cm
- (a) \(VA = \sqrt{8^2 + 34} = \sqrt{98} = 9.90\) cm (3 s.f.)
- (b) \(\tan(VAM) = \dfrac{8}{\sqrt{34}}\), so angle \(VAM = 53.9^\circ\) (1 d.p.)
- \(AC = \sqrt{40^2 + 30^2} = 50\) m
- \(\tan(PCA) = \dfrac{15}{50}\)
- Angle of elevation \(= 16.7^\circ\) (1 d.p.)
- (a) Length \(= \sqrt{5^2 + 4^2 + 3^2} = \sqrt{50} = 7.07\) m (3 s.f.)
- (b) Floor diagonal \(= \sqrt{5^2 + 4^2} = \sqrt{41}\); \(\tan\theta = \dfrac{3}{\sqrt{41}}\), so \(\theta = 25.1^\circ\) (1 d.p.)
- (c) The distance from the string's far end straight back to the end wall is the 5 m length, so \(\sin\phi = \dfrac{5}{\sqrt{50}}\).
- The projection on the wall is the wall's diagonal \(\sqrt{4^2 + 3^2} = 5\) m, so the right-angled triangle has two sides of 5 m: it is isosceles, and \(\phi = 45^\circ\) exactly.
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The diagram shows a cuboid ABCDEFGH with AB = 9 cm, BC = 6 cm and CG = 4 cm. Find the length of the space diagonal AG.
- A11.5 cm
- B10.8 cm
- C19 cm
- D133 cm
For the same cuboid (AB = 9 cm, BC = 6 cm, CG = 4 cm), find the angle \(\theta\) between the diagonal AG and the base ABCD.
- A\(20.3^\circ\)
- B\(24.0^\circ\)
- C\(21.7^\circ\)
- D\(69.7^\circ\)
A pyramid has a square base ABCD of side 10 cm. Its vertex V is directly above the centre M of the base, and each sloping edge is 13 cm. Find the height VM.
- A10.9 cm
- B12 cm
- C8.31 cm
- D14.8 cm
For the same pyramid (square base of side 10 cm, sloping edges 13 cm, vertex above the centre M), find the angle between the sloping edge VA and the base.
- A\(57.0^\circ\)
- B\(67.4^\circ\)
- C\(33.0^\circ\)
- D\(65.4^\circ\)
A rectangular field ABCD has AB = 50 m and BC = 30 m. A vertical flagpole AP of height 20 m stands at corner A. Find the angle of elevation of P from C.
- A\(18.9^\circ\)
- B\(21.8^\circ\)
- C\(33.7^\circ\)
- D\(71.1^\circ\)
A cube has edges of 5 cm. Find the angle between a space diagonal of the cube and its base.
- A\(35.3^\circ\)
- B\(45.0^\circ\)
- C\(54.7^\circ\)
- D\(30.0^\circ\)
Quiz key
\(AC^2 = 9^2 + 6^2 = 117\). Triangle ACG has a right angle at C, so \(AG^2 = 117 + 4^2 = 133\) and \(AG = \sqrt{133} = 11.5\) cm (the same as \(\sqrt{9^2 + 6^2 + 4^2}\)). The tempting 10.8 cm is AC, the diagonal of the base only. 19 cm adds the three edges, and 133 cm forgets the square root.
\(AC = \sqrt{117} = 10.817\) cm, and in triangle ACG, \(\tan\theta = \dfrac{4}{10.817}\), so \(\theta = 20.3^\circ\). The tempting \(24.0^\circ\) uses \(\tan^{-1}\dfrac{4}{9}\), as if AB were the projection of AG; that is the angle in the wrong triangle. \(21.7^\circ\) uses \(\sin^{-1}\) with AC as the hypotenuse, and \(69.7^\circ\) is the angle between AG and the VERTICAL.
\(AC = \sqrt{10^2 + 10^2} = \sqrt{200}\), so \(AM = \tfrac{1}{2}\sqrt{200} = \sqrt{50} = 7.071\) cm. In triangle AMV, \(VM = \sqrt{13^2 - 50} = \sqrt{119} = 10.9\) cm. The tempting 12 cm uses half the SIDE (5 cm) instead of half the diagonal; that gives the height of a sloping face, not of the pyramid. 8.31 cm subtracts the whole side, and 14.8 cm adds the squares.
The angle is angle VAM. \(\cos(\text{VAM}) = \dfrac{AM}{VA} = \dfrac{\sqrt{50}}{13} = 0.5439\), so the angle is \(57.0^\circ\). The tempting \(67.4^\circ\) uses \(\cos^{-1}\dfrac{5}{13}\), half the side instead of half the diagonal; that is the angle between VA and the base EDGE AB. \(33.0^\circ\) is the angle at V, and \(65.4^\circ\) uses \(\tan^{-1}\dfrac{10.9}{5}\), the slope of a face.
\(AC = \sqrt{50^2 + 30^2} = \sqrt{3400} = 58.31\) m. In triangle PAC, with the right angle at A, \(\tan(\text{PCA}) = \dfrac{20}{58.31}\), so the angle of elevation is \(18.9^\circ\). The tempting \(21.8^\circ\) uses the 50 m side, which is the angle seen from B, not C. \(33.7^\circ\) uses 30 m, and \(71.1^\circ\) inverts the fraction.
The face diagonal is \(5\sqrt{2} = 7.071\) cm and the vertical edge is 5 cm, so \(\tan\theta = \dfrac{5}{5\sqrt{2}} = \dfrac{1}{\sqrt{2}}\) and \(\theta = 35.3^\circ\). The tempting \(45.0^\circ\) is the angle between a FACE diagonal and an edge, using \(\tan^{-1}\dfrac{5}{5}\). \(54.7^\circ\) is the angle with the vertical edge, and \(30.0^\circ\) uses the space diagonal \(5\sqrt{3}\) as the adjacent side.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.