The shape of the lesson
By the end of the lesson, learners can
- What is a radian?: Imagine bending a piece of string the length of the radius around the edge of a circle.
- Converting both ways: Every conversion is a multiplication by \(\dfrac{\pi}{180}\) or \(\dfrac{180}{\pi}\).
- Radians in triangles: The triangle formulae you already know work unchanged when the angles are in radians, provided the calculator is in radian mode.
- To radians: multiply by \(\dfrac{\pi}{180}\); leave the answer as a multiple of \(\pi\) when it is exact.
- To degrees: multiply by \(\dfrac{180}{\pi}\); the \(\pi\) cancels.
- Calculator: radian mode for any angle without a degree sign.
- Triangles: \(\tfrac{1}{2}ab\sin C\), sine rule and cosine rule as usual; the angles add to \(\pi\).
- Accuracy: lengths to 3 s.f., radians to 3 s.f., degrees to 1 d.p.
One picture every idea returns to
One radian is the angle at the centre of a circle subtended by an arc equal in length to the radius. A full turn is \(2\pi\) radians, so \(\pi \text{ rad} = 180^\circ\): multiply by \(\dfrac{\pi}{180}\) to go from degrees to radians, and by \(\dfrac{180}{\pi}\) to go back.
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, Measuring angles with the radius.. Degrees are a human choice; radians come from the circle itself. 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: What is a radian? | Imagine bending a piece of string the length of the radius around the edge of a circle. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Converting both ways | Every conversion is a multiplication by \(\dfrac{\pi}{180}\) or \(\dfrac{180}{\pi}\). Model the worked example Degrees to radians on the board, one line at a time. Your turn: Convert \(\dfrac{3\pi}{5}\) to degrees and \(20^\circ\) to radians (exact). | Pairs |
| 21 to 30 min | Teach and check: Radians in triangles | The triangle formulae you already know work unchanged when the angles are in radians, provided the calculator is in radian mode. Model the worked example Area and third side on the board, one line at a time. Your turn: An equilateral triangle has sides of 4 cm. Write each angle in radians and find the exact area. | Round Table |
| 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, 22 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 Radians in triangles, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Your calculator must be in the right mode. \(\sin 1.2\) in degree mode gives \(0.0209\) instead of \(0.932\). | If a question gives angles in radians (or uses \(\pi\)), switch to radian mode before pressing anything. |
| Do not multiply \(\dfrac{7\pi}{12}\) by 180 and stop: that gives \(105\pi\), not \(105^\circ\). | The \(\pi\) in the angle cancels with the \(\pi\) in \(\dfrac{180}{\pi}\). |
| Never mix units in an angle sum. | If two angles are \(\dfrac{\pi}{4}\) and \(1.1\), the third is \(\pi - \dfrac{\pi}{4} - 1.1\), not \(180 - \dfrac{\pi}{4} - 1.1\). |
Radians: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
(a) Express \(72^\circ\) in radians, giving your answer in terms of \(\pi\).
(b) Express \(\dfrac{11\pi}{6}\) radians in degrees.
(a) Convert \(1.75\) radians to degrees, correct to 1 decimal place.
(b) Convert \(200^\circ\) to radians, correct to 3 significant figures.
In triangle \(ABC\), \(AB = 6\) cm, \(BC = 11\) cm and angle \(ABC = 2.1\) radians.
Find (a) the area of the triangle, (b) the length \(AC\).
In triangle \(PQR\), angle \(QPR = \dfrac{\pi}{6}\), angle \(PQR = 0.9\) radians and \(QR = 5\) cm.
(a) Find angle \(PRQ\) in radians.
(b) Find the length \(PQ\).
(c) Find the area of the triangle.
Triangle \(OAB\) has \(OA = OB = 8\) cm and angle \(AOB = \theta\) radians, where \(\theta\) is obtuse. The area of the triangle is \(16\sqrt{3}\ \text{cm}^2\).
Show that \(\theta = \dfrac{2\pi}{3}\), and find the exact length of \(AB\).
A clock has a minute hand of length 12 cm and an hour hand of length 8 cm, both pivoted at the centre of the face.
(a) Find, in radians in terms of \(\pi\), the angle between the hands at exactly 2 o'clock.
(b) Find the exact distance between the tips of the hands at 2 o'clock.
(c) Find the distance between the tips at 4 o'clock, correct to 3 significant figures.
Answer key and marking notes
- (a) \(72 \times \dfrac{\pi}{180} = \dfrac{72\pi}{180} = \dfrac{2\pi}{5}\)
- (b) \(\dfrac{11\pi}{6} \times \dfrac{180}{\pi} = 11 \times 30 = 330^\circ\)
- (a) \(1.75 \times \dfrac{180}{\pi} = 100.267\ldots = 100.3^\circ\)
- (b) \(200 \times \dfrac{\pi}{180} = \dfrac{10\pi}{9} = 3.4906\ldots = 3.49\) rad
- (a) Area \(= \tfrac{1}{2} \times 6 \times 11 \times \sin 2.1 = 33 \times 0.86321 = 28.5\ \text{cm}^2\)
- (b) \(AC^2 = 36 + 121 - 132\cos 2.1 = 157 - 132 \times (-0.50485) = 223.64\ldots\)
- \(AC = 14.95\ldots = 15.0\) cm (3 s.f.)
- (a) \(\pi - \dfrac{\pi}{6} - 0.9 = 1.7179\ldots = 1.72\) rad
- (b) \(PQ\) is opposite angle \(R\) and \(QR\) is opposite angle \(P\): \(PQ = \dfrac{5\sin 1.7180}{\sin(\pi/6)} = \dfrac{5 \times 0.98919}{0.5} = 9.89\) cm
- (c) Angle \(Q\) lies between \(PQ\) and \(QR\): area \(= \tfrac{1}{2} \times 9.8919 \times 5 \times \sin 0.9 = 19.4\ \text{cm}^2\)
- Area \(= \tfrac{1}{2} \times 8 \times 8 \sin\theta = 32\sin\theta = 16\sqrt{3}\), so \(\sin\theta = \dfrac{\sqrt{3}}{2}\).
- The solutions between 0 and \(\pi\) are \(\dfrac{\pi}{3}\) and \(\dfrac{2\pi}{3}\); \(\theta\) is obtuse, so \(\theta = \dfrac{2\pi}{3}\).
- \(AB = 2 \times 8\sin\dfrac{\pi}{3} = 16 \times \dfrac{\sqrt{3}}{2} = 8\sqrt{3}\) cm
- (a) Each hour mark is \(\dfrac{2\pi}{12} = \dfrac{\pi}{6}\), so at 2 o'clock the angle is \(2 \times \dfrac{\pi}{6} = \dfrac{\pi}{3}\).
- (b) \(d^2 = 12^2 + 8^2 - 2 \times 12 \times 8\cos\dfrac{\pi}{3} = 208 - 96 = 112\), so \(d = \sqrt{112} = 4\sqrt{7}\) cm.
- (c) At 4 o'clock the angle is \(\dfrac{2\pi}{3}\): \(d^2 = 208 - 192\cos\dfrac{2\pi}{3} = 208 + 96 = 304\).
- \(d = \sqrt{304} = 17.4\) cm (3 s.f.)
8 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.
Convert \(240^\circ\) to radians, giving your answer as an exact multiple of \(\pi\).
- A\(\dfrac{4\pi}{3}\)
- B\(\dfrac{3\pi}{4}\)
- C\(\dfrac{2\pi}{3}\)
- D\(\dfrac{5\pi}{4}\)
Convert 2.3 radians to degrees, giving your answer correct to 1 decimal place.
- A\(131.8^\circ\)
- B\(0.0401^\circ\)
- C\(414.0^\circ\)
- D\(65.9^\circ\)
Which of these angles is the largest?
- A1.6 radians
- B\(91.5^\circ\)
- C\(\dfrac{\pi}{2}\) radians
- D1.59 radians
Triangle \(ABC\) has \(AB = 6\) cm, \(AC = 10\) cm and angle \(BAC = \dfrac{2\pi}{3}\) radians. Find the exact area of the triangle.
- A\(15\sqrt{3}\ \text{cm}^2\)
- B\(30\sqrt{3}\ \text{cm}^2\)
- C\(15\ \text{cm}^2\)
- D\(1.10\ \text{cm}^2\)
In triangle \(PQR\), \(PQ = 9\) cm, \(PR = 7\) cm and angle \(QPR = 1.3\) radians. Find the length \(QR\).
- A9.81 cm
- B2.01 cm
- C12.8 cm
- D10.6 cm
In triangle \(ABC\), angle \(A = 0.9\) radians, angle \(B = 1.2\) radians and \(BC = 8\) cm. Find the length \(AC\).
- A9.52 cm
- B10.7 cm
- C6.72 cm
- D8.82 cm
A triangle has sides of length 5 cm, 6 cm and 9 cm. Find its largest angle, in radians.
- A1.91
- B1.23
- C0.551
- D109.5
The angles of a triangle are in the ratio \(2 : 3 : 7\). Find the largest angle in radians, as an exact multiple of \(\pi\).
- A\(\dfrac{7\pi}{12}\)
- B\(\dfrac{7\pi}{6}\)
- C\(\dfrac{7}{12}\)
- D\(\dfrac{\pi}{4}\)
Quiz key
Multiply by \(\dfrac{\pi}{180}\): \(240 \times \dfrac{\pi}{180} = \dfrac{4\pi}{3}\). The tempting \(\dfrac{3\pi}{4}\) turns the fraction \(\dfrac{240}{180} = \dfrac{4}{3}\) upside down (it is \(135^\circ\)). \(\dfrac{2\pi}{3}\) is \(120^\circ\) and \(\dfrac{5\pi}{4}\) is \(225^\circ\).
\(2.3 \times \dfrac{180}{\pi} = 131.780\ldots = 131.8^\circ\). The tempting \(0.0401^\circ\) multiplies by \(\dfrac{\pi}{180}\), which is the degrees-to-radians factor. \(414.0^\circ\) multiplies by 180 and forgets to divide by \(\pi\), and \(65.9^\circ\) uses \(\dfrac{90}{\pi}\).
\(1.6 \times \dfrac{180}{\pi} = 91.67^\circ\), \(1.59 \times \dfrac{180}{\pi} = 91.10^\circ\) and \(\dfrac{\pi}{2} = 90^\circ\). So 1.6 radians is the largest, just above \(91.5^\circ\). The tempting \(91.5^\circ\) looks big because 91.5 is a large number, but the other angles are in radians, and a radian is about \(57.3^\circ\).
\(\sin\dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}\), so area \(= \tfrac{1}{2} \times 6 \times 10 \times \dfrac{\sqrt{3}}{2} = 15\sqrt{3}\ \text{cm}^2\). The tempting \(30\sqrt{3}\) leaves out the \(\tfrac{1}{2}\). \(15\) uses \(\tfrac{1}{2}\) for the sine (that is \(\cos\dfrac{\pi}{3}\)), and \(1.10\) comes from a calculator in degree mode, which reads \(\dfrac{2\pi}{3}\) as about \(2.09^\circ\).
\(QR^2 = 9^2 + 7^2 - 2 \times 9 \times 7\cos 1.3 = 130 - 126\cos 1.3 = 96.30\ldots\), so \(QR = 9.81\) cm. The tempting 2.01 cm comes from degree mode (\(\cos 1.3^\circ \approx 1\), so almost everything cancels). 12.8 cm adds \(126\cos 1.3\) instead of subtracting, and 10.6 cm uses \(9 \times 7\cos 1.3\) without the factor 2.
\(\dfrac{AC}{\sin 1.2} = \dfrac{8}{\sin 0.9}\), so \(AC = \dfrac{8\sin 1.2}{\sin 0.9} = 9.52\) cm. The tempting 10.7 cm is the same working in degree mode. 6.72 cm swaps the two sines, and 8.82 cm is \(AB\), the side opposite angle \(C = \pi - 2.1 = 1.04\) radians.
Opposite the 9 cm side: \(\cos C = \dfrac{5^2 + 6^2 - 9^2}{2 \times 5 \times 6} = -\dfrac{1}{3}\), so \(C = \cos^{-1}\left(-\tfrac{1}{3}\right) = 1.91\) radians. The tempting 1.23 drops the minus sign; a negative cosine means an obtuse angle. 0.551 is the smallest angle (opposite the 5 cm side), and 109.5 is the same angle measured in degrees, not radians.
There are \(2 + 3 + 7 = 12\) parts sharing \(\pi\), so one part is \(\dfrac{\pi}{12}\) and the largest angle is \(\dfrac{7\pi}{12}\) (that is \(105^\circ\)). The tempting \(\dfrac{7\pi}{6}\) shares \(2\pi\), which is the angle sum of a full turn, not a triangle. \(\dfrac{7}{12}\) forgets the \(\pi\), and \(\dfrac{\pi}{4}\) is the middle angle.
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.