The shape of the lesson
By the end of the lesson, learners can
- Two sides and the included angle: In the lab, set two sides and open or close the angle between them.
- Three sides, find an angle: Rearranged, the cosine rule gives any angle from the three sides.
- Sine rule or cosine rule?: Look at what you know.
- Side: \(a^2 = b^2 + c^2 - 2bc\cos A\), with \(A\) between \(b\) and \(c\).
- Angle: \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\), subtracting the side opposite \(A\).
- Obtuse: a negative cosine means an obtuse angle; no ambiguity.
- Choose: complete opposite pair, sine rule; two sides and included angle or three sides, cosine rule.
- Accuracy: keep full values; sides 3 s.f., angles 1 d.p.
One picture every idea returns to
\(a^2 = b^2 + c^2 - 2bc\cos A\), where \(A\) is the angle between sides \(b\) and \(c\). To find an angle from three sides, rearrange: \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\). A negative cosine means an obtuse angle.
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, Pythagoras with a correction.. 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: Two sides and the included angle | In the lab, set two sides and open or close the angle between them. Project tab 2 of the lab (Sine and cosine rules lab, cosine rule) and let a student drive it while the class predicts. Model the worked example Finding the third side on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Three sides, find an angle | Rearranged, the cosine rule gives any angle from the three sides. Model the worked example The largest angle on the board, one line at a time. Your turn: Without a calculator, show that the angle opposite the 6 cm side of a triangle with sides 4 cm, 5 cm and 6 cm has cosine \(\dfrac{1}{8}\). | Pairs |
| 21 to 30 min | Teach and check: Sine rule or cosine rule? | Look at what you know. Model the worked example A journey with a turn on the board, one line at a time. | Numbered Heads Together |
| 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, 21 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 Sine rule or cosine rule?, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Typing \(b^2 + c^2 - 2bc\) first and then multiplying by \(\cos A\). | For \(p = 8\), \(q = 11\), \(R = 72^\circ\) that gives \(\sqrt{9\cos 72^\circ} = 1.67\) instead of 11.4. The \(\cos A\) multiplies only the \(2bc\) term. |
| Subtracting the wrong side. | The side you subtract must be the one OPPOSITE the angle you want. Subtracting 5 instead of 9 above gives the smallest angle, not the largest. |
| Using the rounded \(a = 10.4\) in the sine rule. | Keep \(10.436\ldots\) in the calculator, or the angle may come out wrong in the first decimal place. |
The cosine rule: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
In triangle \(ABC\), \(b = 6.2\) cm, \(c = 8.5\) cm and \(A = 37^\circ\). Calculate \(a\).
A triangle has sides of 7 cm, 8 cm and 12 cm. Calculate the size of its largest angle.
In triangle \(ABC\), \(b = 4\) cm, \(c = 9\) cm and \(A = 135^\circ\). Calculate \(a\).
In triangle \(ABC\), \(b = 10\) cm, \(c = 13\) cm and \(A = 52^\circ\).
(a) Calculate \(a\).
(b) Calculate angle \(B\).
A triangle has sides 4 cm, 5 cm and 6 cm. Without a calculator, show that the cosine of its largest angle is \(\dfrac{1}{8}\), and explain why the triangle is acute-angled.
Two hikers leave a camp \(C\) at the same time. Asha walks 6 km on a bearing of \(030^\circ\) to point \(A\). Ben walks 8 km on a bearing of \(110^\circ\) to point \(B\).
(a) Calculate the distance \(AB\).
(b) Calculate angle \(CAB\).
(c) Find the bearing of \(B\) from \(A\).
Answer key and marking notes
- \(a^2 = 6.2^2 + 8.5^2 - 2 \times 6.2 \times 8.5 \times \cos 37^\circ\)
- \(a^2 = 38.44 + 72.25 - 84.177\ldots = 26.512\ldots\)
- \(a = 5.15\) cm (3 s.f.)
- The largest angle faces the 12 cm side.
- \(\cos\theta = \dfrac{7^2 + 8^2 - 12^2}{2 \times 7 \times 8} = \dfrac{-31}{112}\)
- \(\theta = \cos^{-1}(-0.2767\ldots) = 106.1^\circ\) (1 d.p.)
- \(a^2 = 16 + 81 - 72\cos 135^\circ\)
- \(\cos 135^\circ\) is negative, so \(a^2 = 97 + 50.91\ldots = 147.91\ldots\)
- \(a = 12.2\) cm (3 s.f.)
- (a) \(a^2 = 100 + 169 - 260\cos 52^\circ = 108.93\ldots\), so \(a = 10.4\) cm (3 s.f.).
- (b) Sine rule with the pair \(a, A\): \(\sin B = \dfrac{10\sin 52^\circ}{10.436\ldots} = 0.7550\ldots\)
- \(B = 49.0^\circ\) (1 d.p.); acute, since \(b\) is not the longest side.
- The largest angle faces the 6 cm side: \(\cos\theta = \dfrac{4^2 + 5^2 - 6^2}{2 \times 4 \times 5} = \dfrac{5}{40} = \dfrac{1}{8}\).
- \(\dfrac{1}{8} > 0\), so the largest angle is acute.
- The other two angles are smaller still, so every angle is acute.
- (a) Angle \(ACB = 110^\circ - 30^\circ = 80^\circ\). \(AB^2 = 36 + 64 - 96\cos 80^\circ = 83.33\ldots\), so \(AB = 9.13\) km (3 s.f.).
- (b) Sine rule: \(\sin(CAB) = \dfrac{8\sin 80^\circ}{9.128\ldots} = 0.8630\ldots\), so angle \(CAB = 59.7^\circ\) (acute, since \(CB = 8\) km is not the longest side).
- (c) The bearing of \(C\) from \(A\) is \(030^\circ + 180^\circ = 210^\circ\). \(B\) is \(59.7^\circ\) anticlockwise from that direction.
- Bearing of \(B\) from \(A\) \(= 210^\circ - 59.7^\circ = 150.3^\circ\)
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.
In triangle ABC, AC = 8 cm, AB = 11 cm and angle A \(= 47^\circ\). Find the length \(a\) of BC.
- A8.06 cm
- B2.48 cm
- C17.5 cm
- D13.6 cm
In triangle ABC, \(b = 6\) cm, \(c = 9\) cm and angle \(A = 125^\circ\). Find \(a\).
- A13.4 cm
- B7.42 cm
- C10.8 cm
- D179 cm
A triangle has sides 7 cm, 10 cm and 13 cm. Find its largest angle.
- A\(98.2^\circ\)
- B\(32.2^\circ\)
- C\(81.8^\circ\)
- D\(49.6^\circ\)
A boat sails 10 km on a bearing of \(040^\circ\), then 7 km on a bearing of \(160^\circ\). How far is it from its starting point?
- A8.89 km
- B14.8 km
- C12.2 km
- D17 km
A parallelogram has sides 6 cm and 9 cm, and one of its angles is \(70^\circ\). Find the length of the SHORTER diagonal.
- A8.95 cm
- B12.4 cm
- C10.8 cm
- D1.75 cm
In triangle ABC, \(b = 9\) cm, \(c = 12\) cm and angle \(A = 58^\circ\). Find angle \(B\).
- A\(46.5^\circ\)
- B\(46.6^\circ\)
- C\(75.5^\circ\)
- D\(39.5^\circ\)
Quiz key
\(a^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos 47^\circ = 185 - 120.03 = 64.97\), so \(a = 8.06\) cm. The tempting 2.48 cm works out \(64 + 121 - 176\) first and THEN multiplies by \(\cos 47^\circ\). 17.5 cm adds the \(2bc\cos A\) term, and 13.6 cm is \(\sqrt{185}\), Pythagoras with no correction for the angle.
\(a^2 = 36 + 81 - 108\cos 125^\circ = 117 + 61.95 = 178.95\), so \(a = 13.4\) cm. The tempting 7.42 cm drops the negative sign of \(\cos 125^\circ\) and subtracts 61.95. 10.8 cm is \(\sqrt{117}\), ignoring the angle, and 179 cm forgets the square root. Check: the side opposite an obtuse angle must be the longest, and 13.4 cm is.
The largest angle is opposite 13 cm: \(\cos C = \dfrac{7^2 + 10^2 - 13^2}{2 \times 7 \times 10} = \dfrac{-20}{140}\), so \(C = 98.2^\circ\). The negative cosine shows it is obtuse. The tempting \(32.2^\circ\) subtracts \(7^2\), giving the SMALLEST angle. \(81.8^\circ\) drops the minus sign (\(\cos^{-1}\dfrac{20}{140}\)), and \(49.6^\circ\) is the middle angle, opposite 10 cm.
At the turning point, the direction back to the start is \(040^\circ + 180^\circ = 220^\circ\) and the new direction is \(160^\circ\), so the angle inside the triangle is \(220^\circ - 160^\circ = 60^\circ\). Then \(d^2 = 10^2 + 7^2 - 2 \times 10 \times 7 \times \cos 60^\circ = 149 - 70 = 79\), so \(d = 8.89\) km. The tempting 14.8 km uses \(160^\circ - 40^\circ = 120^\circ\), the turn angle rather than the angle inside the triangle. 12.2 km is \(\sqrt{149}\), as if there were a right angle, and 17 km adds the two legs.
The shorter diagonal faces the \(70^\circ\) angle: \(d^2 = 6^2 + 9^2 - 2 \times 6 \times 9 \times \cos 70^\circ = 117 - 36.94 = 80.06\), so \(d = 8.95\) cm. The tempting 12.4 cm is the LONGER diagonal, opposite the \(110^\circ\) angle. 10.8 cm ignores the angle (\(\sqrt{117}\)), and 1.75 cm multiplies \((117 - 108)\) by \(\cos 70^\circ\).
Cosine rule: \(a^2 = 81 + 144 - 216\cos 58^\circ = 110.54\), so \(a = 10.5137\ldots\) cm. Sine rule: \(\sin B = \dfrac{9\sin 58^\circ}{10.5137\ldots} = 0.7259\), so \(B = 46.5^\circ\). The tempting \(46.6^\circ\) comes from rounding \(a\) to 10.5 before the sine rule. \(75.5^\circ\) is angle C, and \(39.5^\circ\) divides by \(c = 12\) instead of \(a\).
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.