The shape of the lesson
By the end of the lesson, learners can
- Where the values come from: In the lab, slide the angle to \(30^\circ\), \(45^\circ\) and \(60^\circ\) and compare each decimal with its exact form: \(0.7071\ldots\) is \(\dfrac{1}{\sqrt{2}}\) and \(0.8660\ldots\) is \(\dfrac{\sqrt{3}}{2}\).
- The table of exact values: Here is everything the syllabus asks for in one place.
- Exact sides without a calculator: When a right-angled triangle has a \(30^\circ\), \(45^\circ\) or \(60^\circ\) angle, SOH CAH TOA gives exact lengths in surd form.
- \(45^\circ\): half a square, sides \(1, 1, \sqrt{2}\).
- \(30^\circ\) and \(60^\circ\): half an equilateral triangle, sides \(1, \sqrt{3}, 2\).
- \(0^\circ\) and \(90^\circ\): sin goes \(0 \to 1\), cos goes \(1 \to 0\).
- Squares: \(\sin^2 x = (\sin x)^2\); square top and bottom.
- Surds: \(\sqrt{3} \times \sqrt{3} = 3\); rationalise if asked.
One picture every idea returns to
Half a square gives sides \(1, 1, \sqrt{2}\) and the \(45^\circ\) values. Half an equilateral triangle of side 2 gives sides \(1, \sqrt{3}, 2\) and the \(30^\circ\) and \(60^\circ\) values. Read off SOH CAH TOA from these triangles.
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, Two triangles, every value.. 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: Where the values come from | In the lab, slide the angle to \(30^\circ\), \(45^\circ\) and \(60^\circ\) and compare each decimal with its exact form: \(0.7071\ldots\) is \(\dfrac{1}{\sqrt{2}}\) and \(0.8660\ldots\) is \(\dfrac{\sqrt{3}}{2}\). Project tab 2 of the lab (Trigonometric graphs lab, exact values) and let a student drive it while the class predicts. Model the worked example Reading values from a triangle on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The table of exact values | Here is everything the syllabus asks for in one place. Model the worked example Evaluating an expression on the board, one line at a time. Your turn: Without a calculator, find the exact value of \(\sin 30^\circ + \cos 60^\circ + \tan 45^\circ\). | Pairs |
| 21 to 30 min | Teach and check: Exact sides without a calculator | When a right-angled triangle has a \(30^\circ\), \(45^\circ\) or \(60^\circ\) angle, SOH CAH TOA gives exact lengths in surd form. Model the worked example Exact sides of a 30-60-90 triangle 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, 17 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 Exact sides without a calculator, 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 |
|---|---|
| Mixing up \(\sin 30^\circ\) and \(\sin 60^\circ\). | Sine grows with the angle, so the smaller angle has the smaller value: \(\sin 30^\circ = \dfrac{1}{2} < \sin 60^\circ = \dfrac{\sqrt{3}}{2}\). |
| \(\sin^2 60^\circ\) means \((\sin 60^\circ)^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}\). | It is not \(\sin(60^2)\) and not \(\dfrac{\sqrt{3}}{4}\): square the top AND the bottom. |
| Dividing by a fraction: \(\dfrac{6}{\frac{1}{2}} = 12\), not 3. | Dividing by \(\dfrac{1}{2}\) is the same as multiplying by 2. |
Exact trigonometric values: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Write down the exact value of (a) \(\sin 45^\circ\) (b) \(\cos 60^\circ\) (c) \(\tan 30^\circ\).
Without a calculator, show that \(\sin^2 60^\circ + \cos^2 60^\circ = 1\).
Without a calculator, find the exact value of \(4\sin 30^\circ + \sqrt{3}\tan 60^\circ\).
In triangle \(ABC\), angle \(B = 90^\circ\), angle \(A = 30^\circ\) and \(AC = 12\) cm.
Without a calculator, find (a) \(BC\) (b) \(AB\), giving your answer in surd form (c) the exact area of the triangle.
Without a calculator, find the acute angle \(x\) in each case. (a) \(2\sin x = 1\) (b) \(\sqrt{3}\tan x = 3\)
A regular hexagon is made of six equilateral triangles, each of side 10 cm.
Without a calculator, find (a) the exact height of one triangle (b) the exact area of the hexagon.
Answer key and marking notes
- (a) \(\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\)
- (b) \(\dfrac{1}{2}\)
- (c) \(\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\)
- \(\sin^2 60^\circ = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}\) and \(\cos^2 60^\circ = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}\).
- \(\dfrac{3}{4} + \dfrac{1}{4} = 1\), as required.
- \(4 \times \dfrac{1}{2} = 2\) and \(\sqrt{3} \times \sqrt{3} = 3\).
- \(2 + 3 = 5\)
- (a) \(BC = 12\sin 30^\circ = 6\) cm
- (b) \(AB = 12\cos 30^\circ = 12 \times \dfrac{\sqrt{3}}{2} = 6\sqrt{3}\) cm
- (c) Area \(= \tfrac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\) cm2
- (a) \(\sin x = \dfrac{1}{2}\), so \(x = 30^\circ\).
- (b) \(\tan x = \dfrac{3}{\sqrt{3}} = \sqrt{3}\), so \(x = 60^\circ\).
- (a) Each angle of an equilateral triangle is \(60^\circ\), so height \(= 10\sin 60^\circ = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}\) cm.
- (b) Area of one triangle \(= \tfrac{1}{2} \times 10 \times 5\sqrt{3} = 25\sqrt{3}\) cm2.
- Hexagon \(= 6 \times 25\sqrt{3} = 150\sqrt{3}\) cm2.
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.
Write down the exact value of \(\tan 30^\circ\).
- A\(\dfrac{\sqrt{3}}{3}\)
- B\(\sqrt{3}\)
- C\(\dfrac{1}{2}\)
- D\(\dfrac{\sqrt{3}}{2}\)
Without a calculator, find the exact value of \(\tan 45^\circ + \sin^2 60^\circ\).
- A\(\dfrac{7}{4}\)
- B\(1 + \dfrac{\sqrt{3}}{4}\)
- C\(\dfrac{5}{2}\)
- D\(\dfrac{3}{4}\)
Without a calculator, find the acute angle \(x\) for which \(2\sin x = \sqrt{2}\).
- A\(45^\circ\)
- B\(30^\circ\)
- C\(60^\circ\)
- D\(135^\circ\)
Triangle ABC has a right angle at B, AC = 14 cm and angle BAC \(= 60^\circ\). Without a calculator, find the exact length \(x\) of BC.
- A\(7\sqrt{3}\) cm
- B7 cm
- C\(14\sqrt{3}\) cm
- D\(\dfrac{28\sqrt{3}}{3}\) cm
A right-angled triangle has an angle of \(30^\circ\). The side adjacent to this angle is 5 cm. Without a calculator, find the exact length of the hypotenuse.
- A\(\dfrac{10\sqrt{3}}{3}\) cm
- B\(\dfrac{5\sqrt{3}}{2}\) cm
- C10 cm
- D\(\dfrac{5\sqrt{3}}{3}\) cm
Without a calculator, find the exact area of an equilateral triangle of side 8 cm.
- A\(16\sqrt{3}\) cm2
- B32 cm2
- C\(32\sqrt{3}\) cm2
- D\(8\sqrt{3}\) cm2
Quiz key
In the \(1, \sqrt{3}, 2\) triangle, the side opposite \(30^\circ\) is 1 and the adjacent side is \(\sqrt{3}\), so \(\tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\). The tempting \(\sqrt{3}\) is \(\tan 60^\circ\), the fraction upside down. \(\dfrac{1}{2}\) is \(\sin 30^\circ\), and \(\dfrac{\sqrt{3}}{2}\) is \(\cos 30^\circ\).
\(\tan 45^\circ = 1\) and \(\sin^2 60^\circ = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}\), so the total is \(1 + \dfrac{3}{4} = \dfrac{7}{4}\). The tempting \(1 + \dfrac{\sqrt{3}}{4}\) squares only the bottom. \(\dfrac{5}{2}\) squares only the top (\(\dfrac{3}{2}\)), and \(\dfrac{3}{4}\) takes \(\tan 45^\circ\) as 0.
\(\sin x = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}\), which is \(\sin 45^\circ\) (half a square, sides \(1, 1, \sqrt{2}\)), so \(x = 45^\circ\). The tempting \(30^\circ\) matches \(\sin x = \dfrac{1}{2}\), ignoring the \(\sqrt{2}\). \(60^\circ\) has \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), and \(135^\circ\) also has this sine but is not acute.
\(x = 14\sin 60^\circ = 14 \times \dfrac{\sqrt{3}}{2} = 7\sqrt{3}\) cm. The tempting 7 cm uses \(\cos 60^\circ = \dfrac{1}{2}\), which gives AB, the side ADJACENT to the angle. \(14\sqrt{3}\) cm forgets to halve, and \(\dfrac{28\sqrt{3}}{3} = \dfrac{14}{\sin 60^\circ}\) divides instead of multiplying, giving a side longer than the hypotenuse.
\(h = \dfrac{5}{\cos 30^\circ} = \dfrac{5}{\frac{\sqrt{3}}{2}} = \dfrac{10}{\sqrt{3}} = \dfrac{10\sqrt{3}}{3}\) cm. The tempting \(\dfrac{5\sqrt{3}}{2}\) multiplies by \(\cos 30^\circ\) instead of dividing; it is shorter than the 5 cm side. 10 cm divides by \(\sin 30^\circ\), and \(\dfrac{5\sqrt{3}}{3} = 5\tan 30^\circ\) is the opposite side.
Height \(= 8\sin 60^\circ = 8 \times \dfrac{\sqrt{3}}{2} = 4\sqrt{3}\) cm, so area \(= \tfrac{1}{2} \times 8 \times 4\sqrt{3} = 16\sqrt{3}\) cm2. The tempting 32 cm2 uses the sloping side 8 cm as the height. \(32\sqrt{3}\) forgets the half, and \(8\sqrt{3}\) uses half the base AND the half.
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.