The shape of the lesson
By the end of the lesson, learners can
- Opposite pairs: In the lab, change the angles of the triangle and watch the three ratios \(\dfrac{a}{\sin A}\), \(\dfrac{b}{\sin B}\), \(\dfrac{c}{\sin C}\).
- Obtuse angles and the ambiguous case: Because \(\sin(180^\circ - x) = \sin x\), the equation \(\sin B = 0.86\) has two answers between \(0^\circ\) and \(180^\circ\): an acute one and an obtuse one.
- The sine rule with bearings: Bearings problems often give two angles and one distance, which is exactly the sine rule's job.
- Label: side \(a\) opposite angle \(A\), \(b\) opposite \(B\), \(c\) opposite \(C\).
- Pair: find a side and its opposite angle (use \(180^\circ\) for the third angle if needed).
- Side: \(b = \dfrac{a\sin B}{\sin A}\).
- Angle: \(\sin B = \dfrac{b\sin A}{a}\); then ask whether \(180^\circ - B\) also works.
- Accuracy: sides 3 s.f., angles 1 d.p.; keep full values between steps.
One picture every idea returns to
In any triangle \(ABC\), with side \(a\) opposite angle \(A\) and so on: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\). Use it when you know a side AND its opposite angle. To find an angle, flip it: \(\dfrac{\sin A}{a} = \dfrac{\sin B}{b}\).
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, Any triangle, no right angle needed.. 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: Opposite pairs | In the lab, change the angles of the triangle and watch the three ratios \(\dfrac{a}{\sin A}\), \(\dfrac{b}{\sin B}\), \(\dfrac{c}{\sin C}\). Project tab 1 of the lab (Sine and cosine rules lab, sine rule) and let a student drive it while the class predicts. Model the worked example Finding a side on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Obtuse angles and the ambiguous case | Because \(\sin(180^\circ - x) = \sin x\), the equation \(\sin B = 0.86\) has two answers between \(0^\circ\) and \(180^\circ\): an acute one and an obtuse one. Model the worked example Two possible triangles on the board, one line at a time. Your turn: Explain why no triangle exists with \(A = 50^\circ\), \(a = 4\) cm and \(b = 7\) cm. | Pairs |
| 21 to 30 min | Teach and check: The sine rule with bearings | Bearings problems often give two angles and one distance, which is exactly the sine rule's job. Model the worked example Locating a ship 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, 18 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 The sine rule with bearings, 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 |
|---|---|
| Pairing a side with an angle that is NOT opposite it. | Write the triangle's letters on your sketch and check: side \(a\) must face angle \(A\). Using \(\sin 73^\circ\) where the opposite angle is really \(55^\circ\) is the commonest lost mark. |
| Always giving the calculator's acute angle. | If the question says the angle is obtuse, or your diagram shows it is, the answer is \(180^\circ\) minus the calculator value. |
| Putting the bearing itself (\(040^\circ\) or \(300^\circ\)) into the sine rule. | Bearings are measured from north; the sine rule needs the angles inside the triangle. |
The sine rule: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
In triangle \(ABC\), \(A = 38^\circ\), \(B = 71^\circ\) and \(a = 6.4\) cm. Calculate \(b\).
In triangle \(ABC\), \(b = 8.3\) cm, \(c = 6.1\) cm and the obtuse angle \(B = 112^\circ\). Calculate angle \(C\).
In triangle \(ABC\), \(A = 55^\circ\), \(B = 48^\circ\) and \(c = 14\) cm. Calculate \(a\).
In triangle \(ABC\), \(A = 40^\circ\), \(a = 7\) cm and \(b = 10\) cm.
(a) Show that there are two possible values for angle \(B\), and find them.
(b) Find the two possible values of angle \(C\).
Show that it is not possible to draw a triangle \(ABC\) with \(A = 50^\circ\), \(a = 4\) cm and \(b = 7\) cm.
Two points \(A\) and \(B\) are 50 m apart on one straight bank of a river. A tree \(T\) stands on the opposite bank. Angle \(TAB = 62^\circ\) and angle \(TBA = 75^\circ\).
(a) Calculate the distance \(AT\).
(b) Calculate the width of the river (the perpendicular distance from \(T\) to the line \(AB\)).
Answer key and marking notes
- \(\dfrac{b}{\sin 71^\circ} = \dfrac{6.4}{\sin 38^\circ}\)
- \(b = \dfrac{6.4\sin 71^\circ}{\sin 38^\circ} = 9.83\) cm (3 s.f.)
- \(\sin C = \dfrac{6.1\sin 112^\circ}{8.3} = 0.6814\ldots\)
- \(C = 43.0^\circ\) (1 d.p.)
- \(C\) cannot be \(180^\circ - 43.0^\circ = 137.0^\circ\), since \(B\) is already obtuse.
- \(C = 180^\circ - 55^\circ - 48^\circ = 77^\circ\), which pairs with \(c = 14\).
- \(a = \dfrac{14\sin 55^\circ}{\sin 77^\circ} = 11.8\) cm (3 s.f.)
- (a) \(\sin B = \dfrac{10\sin 40^\circ}{7} = 0.9183\ldots\), so \(B = 66.7^\circ\) or \(B = 180^\circ - 66.7^\circ = 113.3^\circ\).
- Both are possible because \(40^\circ + 113.3^\circ = 153.3^\circ < 180^\circ\).
- (b) \(C = 180^\circ - 40^\circ - 66.7^\circ = 73.3^\circ\) or \(C = 180^\circ - 40^\circ - 113.3^\circ = 26.7^\circ\).
- Using the sine rule, \(\sin B = \dfrac{7\sin 50^\circ}{4} = 1.34\) (3 s.f.).
- The sine of an angle cannot exceed 1, so no such angle \(B\) exists, and no such triangle can be drawn.
- (a) Angle \(ATB = 180^\circ - 62^\circ - 75^\circ = 43^\circ\), which faces \(AB = 50\) m.
- \(AT\) faces the \(75^\circ\) angle: \(AT = \dfrac{50\sin 75^\circ}{\sin 43^\circ} = 70.8\) m (3 s.f.)
- (b) In the right-angled triangle from \(T\) to the bank, \(AT\) is the hypotenuse and the width is opposite the \(62^\circ\):
- width \(= 70.815\ldots \times \sin 62^\circ = 62.5\) m (3 s.f.)
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, angle A \(= 42^\circ\), angle B \(= 71^\circ\) and BC = 9 cm. Find the length \(b\) of AC.
- A12.7 cm
- B6.37 cm
- C12.4 cm
- D8.51 cm
In triangle ABC, \(a = 11\) cm, \(b = 8\) cm and angle \(A = 64^\circ\). Find angle \(B\).
- A\(40.8^\circ\)
- B\(139.2^\circ\)
- C\(46.5^\circ\)
- D\(49.2^\circ\)
In triangle ABC, angle \(A = 32^\circ\), \(a = 7\) cm and \(b = 11\) cm. Find all the possible values of angle \(B\).
- A\(56.4^\circ\) and \(123.6^\circ\)
- B\(56.4^\circ\) only
- C\(123.6^\circ\) only
- D\(56.4^\circ\) and \(236.4^\circ\)
In triangle ABC, \(a = 10\) cm, \(c = 13\) cm and angle \(A = 40^\circ\). Angle \(C\) is obtuse. Find angle \(C\).
- A\(123.3^\circ\)
- B\(56.7^\circ\)
- C\(16.7^\circ\)
- D\(83.3^\circ\)
Station B is 30 km due east of station A. A ship S is on a bearing of \(052^\circ\) from A and \(320^\circ\) from B. Find the distance AS.
- A23.0 km
- B18.5 km
- C19.3 km
- D37.3 km
In triangle ABC, angle \(A = 50^\circ\) and \(a = 4\) cm. For which length of \(b\) is it impossible to draw the triangle?
- A\(b = 7\) cm
- B\(b = 5\) cm
- C\(b = 4.5\) cm
- D\(b = 3\) cm
Quiz key
\(\dfrac{b}{\sin 71^\circ} = \dfrac{9}{\sin 42^\circ}\), so \(b = \dfrac{9\sin 71^\circ}{\sin 42^\circ} = 12.71\ldots = 12.7\) cm. The tempting 6.37 cm turns the fraction upside down, \(\dfrac{9\sin 42^\circ}{\sin 71^\circ}\). 12.4 cm uses angle C \(= 67^\circ\), which is opposite AB, not AC, and 8.51 cm is \(9\sin 71^\circ\), treating the triangle as right-angled.
\(\sin B = \dfrac{8\sin 64^\circ}{11} = 0.6537\), so \(B = 40.8^\circ\). B cannot be obtuse here: \(b \lt a\), so B must be smaller than A. The tempting \(139.2^\circ\) is \(180^\circ - 40.8^\circ\); with A \(= 64^\circ\) it would make the angle sum over \(180^\circ\). \(46.5^\circ\) scales the ANGLE in proportion to the sides (\(64 \times \dfrac{8}{11}\)), which is not how the rule works, and \(49.2^\circ\) is \(90^\circ - 40.8^\circ\).
\(\sin B = \dfrac{11\sin 32^\circ}{7} = 0.8327\), so \(B = 56.4^\circ\) or \(B = 180^\circ - 56.4^\circ = 123.6^\circ\). Both work, because \(32^\circ + 123.6^\circ \lt 180^\circ\) (and \(b \gt a\), the ambiguous case). The tempting "\(56.4^\circ\) only" trusts the calculator and misses the obtuse triangle. \(236.4^\circ\) adds \(180^\circ\), too big for any triangle.
\(\sin C = \dfrac{13\sin 40^\circ}{10} = 0.8356\). The calculator gives \(56.7^\circ\), but C is obtuse, so \(C = 180^\circ - 56.7^\circ = 123.3^\circ\). The tempting \(56.7^\circ\) ignores the word obtuse. \(16.7^\circ\) is angle B in the obtuse triangle, and \(83.3^\circ\) treats \(56.7^\circ\) as angle B and finds C from the angle sum.
Angle SAB \(= 90^\circ - 52^\circ = 38^\circ\). From B, \(320^\circ\) is \(40^\circ\) west of north, so angle SBA \(= 90^\circ - 40^\circ = 50^\circ\) and angle ASB \(= 180^\circ - 38^\circ - 50^\circ = 92^\circ\). AS is opposite angle B: \(AS = \dfrac{30\sin 50^\circ}{\sin 92^\circ} = 22.99\ldots = 23.0\) km. The tempting 18.5 km is BS, opposite the \(38^\circ\) angle. 19.3 km uses the angles from north (\(52^\circ\) and \(40^\circ\)) as triangle angles, and 37.3 km pairs AS with \(\sin 38^\circ\) instead of the angle at S.
With \(b = 7\): \(\sin B = \dfrac{7\sin 50^\circ}{4} = 1.34\), and no angle has a sine greater than 1, so the triangle cannot exist. For \(b = 5\), \(\sin B = 0.958\); for \(b = 4.5\), \(0.862\); for \(b = 3\), \(0.575\): all possible. The tempting \(b = 3\) looks too short to reach, but a side shorter than \(a\) always gives a triangle; the problem is a side so long that \(a\) cannot reach the third 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.