The shape of the lesson
By the end of the lesson, learners can
- Label, choose, solve: In the lab, pick an angle and watch the three sides relabel themselves as the angle moves to the other corner.
- Finding an angle: When two sides are known, find the ratio, then use the inverse key (\(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\), usually SHIFT then sin, cos or tan) to turn the ratio back into an angle.
- Problems in two dimensions: Real problems often join two right-angled triangles along a common side.
- Label: hypotenuse first (opposite the right angle), then opposite and adjacent from the angle.
- Choose: SOH, CAH or TOA, whichever uses the two sides in the question.
- Unknown on top: multiply. Unknown on the bottom: divide.
- Angle: work out the ratio, then use \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\).
- Accuracy: sides to 3 s.f., angles to 1 d.p.; keep full values between steps.
One picture every idea returns to
Label the sides from the angle: hypotenuse (opposite the right angle), opposite (facing the angle), adjacent (next to the angle). Then \(\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\).
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, Angles into lengths.. Pythagoras links three sides. 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: Label, choose, solve | In the lab, pick an angle and watch the three sides relabel themselves as the angle moves to the other corner. Project tab 3 of the lab (Pythagoras and trigonometry lab, SOH CAH TOA) and let a student drive it while the class predicts. Model the worked example Unknown on top on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Finding an angle | When two sides are known, find the ratio, then use the inverse key (\(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\), usually SHIFT then sin, cos or tan) to turn the ratio back into an angle. Model the worked example Inverse tangent on the board, one line at a time. Your turn: In a right-angled triangle the hypotenuse is 5.8 cm and the side opposite angle \(y\) is 3.2 cm. Find \(y\). | Pairs |
| 21 to 30 min | Teach and check: Problems in two dimensions | Real problems often join two right-angled triangles along a common side. Model the worked example Two triangles, one shared side 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, 16 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 two dimensions, 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 |
|---|---|
| Writing \(h = 6\cos 52^\circ = 3.69\) cm. | When the unknown is on the bottom of the fraction you divide, not multiply. A 'hypotenuse' shorter than the other side is a sure sign of this slip. |
| Using the full base (10 cm) in the isosceles triangle gives \(\cos^{-1}(10/12) = 33.6^\circ\). | The right angle is only in the half-triangle, so use half the base. |
| Rounding \(AD\) too early. | With \(AD = 5.8\) you only just survive (\(DC = 12.44\)); with \(AD = 6\) you get \(DC = 12.9\) cm and lose the accuracy mark. Keep the unrounded value in your calculator. |
Right-angled trigonometry: SOH CAH TOA: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
In triangle \(ABC\), angle \(B = 90^\circ\), \(AC = 12\) cm and angle \(A = 34^\circ\). Calculate \(AB\).
In a right-angled triangle, the side adjacent to an angle of \(48^\circ\) is 7.5 cm. Calculate the length of the hypotenuse.
In a right-angled triangle, the two shorter sides are 11 cm and 6.5 cm. Calculate the angle opposite the 11 cm side.
A building regulation says a wheelchair ramp must make an angle of no more than \(5^\circ\) with the horizontal. A ramp rises 0.5 m over a horizontal distance of 6 m.
Does the ramp meet the regulation? Show your working.
An isosceles triangle has two sides of 8 cm. The angle between them is \(50^\circ\). Calculate the length of the third side.
A vertical mast \(PQ\) is 6 m tall, with \(Q\) on horizontal ground. A wire joins the top \(P\) to a point \(R\) on the ground and makes an angle of \(58^\circ\) with the ground. A second wire joins \(P\) to a point \(S\) on the ground on the other side of the mast, making an angle of \(40^\circ\) with the ground. \(R\), \(Q\) and \(S\) lie in a straight line.
Calculate the distance \(RS\).
Answer key and marking notes
- From angle \(A\), \(AB\) is adjacent and \(AC\) is the hypotenuse, so CAH.
- \(AB = 12\cos 34^\circ = 9.95\) cm (3 s.f.)
- \(\cos 48^\circ = \dfrac{7.5}{h}\)
- \(h = \dfrac{7.5}{\cos 48^\circ} = 11.2\) cm (3 s.f.)
- Opposite 11 cm, adjacent 6.5 cm: \(\tan\theta = \dfrac{11}{6.5}\)
- \(\theta = \tan^{-1}(1.6923\ldots) = 59.4^\circ\) (1 d.p.)
- Opposite 0.5 m, adjacent 6 m: \(\tan\theta = \dfrac{0.5}{6}\)
- \(\theta = \tan^{-1}(0.08333\ldots) = 4.8^\circ\) (1 d.p.)
- \(4.8^\circ \le 5^\circ\), so the ramp meets the regulation.
- The line of symmetry halves the \(50^\circ\) angle and the base, giving a right-angled triangle with hypotenuse 8 cm and angle \(25^\circ\).
- Half the base is opposite \(25^\circ\): \(8\sin 25^\circ = 3.3809\ldots\) cm
- Base \(= 2 \times 3.3809\ldots = 6.76\) cm (3 s.f.)
- Triangle \(PQR\): \(\tan 58^\circ = \dfrac{6}{QR}\), so \(QR = \dfrac{6}{\tan 58^\circ} = 3.749\ldots\) m
- Triangle \(PQS\): \(QS = \dfrac{6}{\tan 40^\circ} = 7.150\ldots\) m
- \(RS = QR + QS = 3.749\ldots + 7.150\ldots = 10.899\ldots\)
- \(RS = 10.9\) 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.
Triangle ABC has a right angle at B. AC = 14 cm and angle BAC \(= 38^\circ\). Find the length \(x\) of BC.
- A8.62 cm
- B11.0 cm
- C10.9 cm
- D22.7 cm
In a right-angled triangle, the side adjacent to an angle of \(41^\circ\) is 9 cm. Find the length of the hypotenuse.
- A11.9 cm
- B6.79 cm
- C13.7 cm
- D7.82 cm
In a right-angled triangle, the side opposite angle \(x\) is 5.2 cm and the side adjacent to \(x\) is 8.4 cm. Find \(x\).
- A\(31.8^\circ\)
- B\(58.2^\circ\)
- C\(38.2^\circ\)
- D\(51.8^\circ\)
A ladder 7 m long leans against a vertical wall and makes an angle of \(72^\circ\) with the horizontal ground. How far up the wall does the ladder reach?
- A6.66 m
- B2.16 m
- C21.5 m
- D7.36 m
An isosceles triangle has two equal sides of 15 cm and a base of 18 cm. Find the size of each base angle.
- A\(53.1^\circ\)
- B\(36.9^\circ\)
- C\(73.7^\circ\)
- D\(31.0^\circ\)
In triangle ABC, AD is perpendicular to BC. AB = 11 cm, angle ABC \(= 35^\circ\) and angle ACB \(= 48^\circ\). Find the length \(x\) of DC.
- A5.68 cm
- B7.01 cm
- C5.67 cm
- D9.01 cm
Quiz key
BC is opposite the \(38^\circ\) angle and AC is the hypotenuse, so use sine: \(x = 14\sin 38^\circ = 8.619\ldots = 8.62\) cm. The tempting 11.0 cm uses cosine, which gives the adjacent side AB. 10.9 cm uses \(14\tan 38^\circ\), and 22.7 cm divides by \(\sin 38^\circ\), giving a side longer than the hypotenuse.
\(\cos 41^\circ = \dfrac{9}{h}\), so \(h = \dfrac{9}{\cos 41^\circ} = 11.92\ldots = 11.9\) cm. The tempting 6.79 cm multiplies, \(9\cos 41^\circ\); a hypotenuse shorter than the 9 cm side is impossible. 13.7 cm divides by \(\sin 41^\circ\), and 7.82 cm is \(9\tan 41^\circ\), the opposite side.
\(\tan x = \dfrac{5.2}{8.4}\), so \(x = \tan^{-1}(0.619\ldots) = 31.8^\circ\) (1 d.p.). The tempting \(58.2^\circ\) inverts the fraction, \(\tan^{-1}\dfrac{8.4}{5.2}\), which gives the OTHER acute angle. \(38.2^\circ\) uses \(\sin^{-1}\) and \(51.8^\circ\) uses \(\cos^{-1}\), treating 8.4 cm as the hypotenuse.
Height \(= 7\sin 72^\circ = 6.657\ldots = 6.66\) m. The tempting 2.16 m uses cosine, which gives the distance of the foot from the wall. 21.5 m uses tangent, and 7.36 m divides by \(\sin 72^\circ\); both are longer than the ladder.
The line of symmetry halves the base, giving a right-angled triangle with hypotenuse 15 cm and adjacent side 9 cm. Base angle \(= \cos^{-1}\dfrac{9}{15} = 53.1^\circ\). The tempting \(36.9^\circ\) uses \(\sin^{-1}\dfrac{9}{15}\), which is half the angle at the top, not the base angle. \(73.7^\circ\) is the whole top angle, and \(31.0^\circ\) uses \(\tan^{-1}\dfrac{9}{15}\), wrongly treating 15 cm as the adjacent side.
In triangle ABD: \(AD = 11\sin 35^\circ = 6.309\ldots\) cm. In triangle ADC, AD is opposite the \(48^\circ\) angle and DC is adjacent: \(x = \dfrac{6.309\ldots}{\tan 48^\circ} = 5.680\ldots = 5.68\) cm. The tempting 7.01 cm multiplies by \(\tan 48^\circ\) instead of dividing. 5.67 cm comes from rounding AD to 6.3 too early, and 9.01 cm is BD, not DC.
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.