The shape of the lesson
By the end of the lesson, learners can
- The angle in a semicircle: In the semicircle tab, drag point C around the circle.
- Tangent meets radius: In the tangents tab, drag the point of contact around the circle and watch the tangent turn with it.
- Putting it together: The find-the-angle tab gives random circle diagrams.
- Diameter? Right angle at the circumference (angle in a semicircle).
- Tangent? Right angle where it meets the radius.
- Radii? Isosceles triangle, equal base angles.
- Finish with the angle sum, Pythagoras or trigonometry.
- Reasons in syllabus words for every angle.
One picture every idea returns to
The angle in a semicircle is \(90^\circ\): a triangle drawn on a diameter with its third corner on the circle has a right angle at that corner. A tangent meets the radius at its point of contact at \(90^\circ\).
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, Hidden right angles.. Two circle facts plant a right angle wherever they apply: the angle in a semicircle, and the angle between a tangent and a radius. 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: The angle in a semicircle | In the semicircle tab, drag point C around the circle. Project tab 2 of the lab (Circle theorems lab, semicircle) and let a student drive it while the class predicts. Model the worked example Using the right angle on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Tangent meets radius | In the tangents tab, drag the point of contact around the circle and watch the tangent turn with it. Project tab 3 of the lab (Circle theorems lab, tangents) and let a student drive it while the class predicts. Model the worked example Angle in the tangent triangle on the board, one line at a time. Your turn: A circle has centre O and radius 8 cm. A point P is 17 cm from O. Find the length of the tangent from P to the circle. | Pairs |
| 21 to 30 min | Teach and check: Putting it together | The find-the-angle tab gives random circle diagrams. Project tab 9 of the lab (Circle theorems lab, find the angle) and let a student drive it while the class predicts. Model the worked example Tangent and diameter together on the board, one line at a time. Your turn: In the same diagram, find angle CBD. (Hint: angle ACB is in a semicircle.) | 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 Putting it together, 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 |
|---|---|
| The right angle is at C, the point on the circumference, NOT at the centre and not at A or B. | It only holds when AB really is a diameter, so check the chord passes through O. |
| OP, the line from the centre to the outside point, is the hypotenuse, not the tangent. | Subtract squares to find the tangent: \(PT^2 = OP^2 - OT^2\). |
| Write the circle reason in the syllabus words. 'Because it touches the circle' does not earn the mark; 'angle between tangent and radius = 90°' does. | Return to Putting it together: ask a pair to state the correct version and show it on the lab before moving on. |
Circle theorems I: semicircles and tangents: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
Complete each statement. (a) The angle in a semicircle is ... (b) The angle between a tangent and a radius at the point of contact is ...
AB is a diameter of a circle and C lies on the circle. Angle BAC \(= 3x\) and angle ABC \(= 2x\). Find \(x\) and angle BAC.
A circle has centre O and radius 8 cm. PT is a tangent to the circle at T and OP = 17 cm. Find PT.
A circle has centre O. AB is a diameter and C is a point on the circle with angle OCB \(= 63^\circ\).
Find (a) angle OBC (b) angle BOC (c) angle OCA. Give a reason for each answer.
AB is a diameter of a circle with centre O, and C is any point on the circle. Let angle OAC \(= a\) and angle OBC \(= b\).
Prove that angle ACB \(= 90^\circ\).
A satellite S is 8000 km from the centre O of a planet of radius 6400 km. The line of sight from S just touches the planet's surface at T.
(a) Explain why angle OTS \(= 90^\circ\). (b) Find the distance ST. (c) Find angle TOS, giving your answer to 1 decimal place.
Answer key and marking notes
- (a) \(90^\circ\)
- (b) \(90^\circ\)
- Angle ACB \(= 90^\circ\) (angle in a semicircle).
- \(3x + 2x + 90 = 180\), so \(5x = 90\) and \(x = 18\).
- Angle BAC \(= 3 \times 18 = 54^\circ\).
- Angle OTP \(= 90^\circ\) (angle between tangent and radius).
- \(PT^2 = 17^2 - 8^2 = 289 - 64 = 225\).
- PT = 15 cm.
- (a) OB = OC (radii), so triangle OBC is isosceles: angle OBC = angle OCB \(= 63^\circ\).
- (b) Angle BOC \(= 180^\circ - 63^\circ - 63^\circ = 54^\circ\) (angles in a triangle).
- (c) Angle ACB \(= 90^\circ\) (angle in a semicircle), so angle OCA \(= 90^\circ - 63^\circ = 27^\circ\).
- OA = OC (radii), so triangle OAC is isosceles and angle OCA \(= a\).
- OB = OC (radii), so triangle OBC is isosceles and angle OCB \(= b\).
- Angle ACB \(= a + b\). The angles of triangle ABC are \(a\), \(b\) and \(a + b\), so \(2a + 2b = 180^\circ\).
- Therefore \(a + b = 90^\circ\), so angle ACB \(= 90^\circ\).
- (a) ST is a tangent to the circle and OT is a radius at the point of contact; the angle between a tangent and a radius is \(90^\circ\).
- (b) \(ST^2 = 8000^2 - 6400^2 = 64\,000\,000 - 40\,960\,000 = 23\,040\,000\).
- \(ST = \sqrt{23\,040\,000} = 4800\) km.
- (c) \(\cos(\text{TOS}) = \dfrac{OT}{OS} = \dfrac{6400}{8000} = 0.8\), so angle TOS \(= 36.9^\circ\).
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.
AB is a diameter of the circle, centre O, and C is a point on the circle. Angle CAB \(= 34^\circ\). Find angle ABC, marked \(x\).
- A\(56^\circ\)
- B\(34^\circ\)
- C\(146^\circ\)
- D\(68^\circ\)
A circle has centre O and radius 5 cm. PT is a tangent to the circle at T, and OP = 13 cm. Find the length of PT.
- A12 cm
- B13.9 cm
- C8 cm
- D18 cm
AB is a diameter of a circle with centre O, and C is a point on the circle. Angle OCA \(= 25^\circ\). Find angle BOC.
- A\(50^\circ\)
- B\(130^\circ\)
- C\(65^\circ\)
- D\(25^\circ\)
PT is a tangent to a circle with centre O, touching the circle at T. Which angle must be \(90^\circ\)?
- AAngle OTP
- BAngle OPT
- CAngle TOP
- DAngle PTO only if OP is a diameter
AB is a diameter of the circle, centre O. The tangent at B meets AC extended at D. Angle CAB \(= 38^\circ\). Find angle CBD, marked \(x\).
- A\(38^\circ\)
- B\(52^\circ\)
- C\(90^\circ\)
- D\(19^\circ\)
P is a point 10 cm from the centre O of a circle. The tangent from P touches the circle at T, and PT = 8 cm. Find the diameter of the circle.
- A12 cm
- B6 cm
- C12.8 cm
- D25.6 cm
Quiz key
The angle in a semicircle is \(90^\circ\), so angle ACB \(= 90^\circ\) and \(x = 180^\circ - 90^\circ - 34^\circ = 56^\circ\). The tempting \(34^\circ\) assumes the triangle is isosceles with equal angles at A and B, but CA and CB are not equal. \(146^\circ\) leaves out the right angle, and \(68^\circ\) is the angle COB at the centre.
The tangent meets the radius at \(90^\circ\), so triangle OTP has its right angle at T and OP is the hypotenuse: \(PT = \sqrt{13^2 - 5^2} = \sqrt{144} = 12\) cm. The tempting 13.9 cm ADDS the squares (\(\sqrt{194}\)), treating PT as the hypotenuse. 8 cm subtracts the lengths, and 18 cm adds them.
OA = OC (radii), so angle OAC = angle OCA \(= 25^\circ\) and angle AOC \(= 180^\circ - 50^\circ = 130^\circ\). AOB is a straight line, so angle BOC \(= 180^\circ - 130^\circ = 50^\circ\). The tempting \(130^\circ\) is angle AOC, the angle on the other side of OC. \(65^\circ\) is angle OCB (from \(90^\circ - 25^\circ\)), and \(25^\circ\) copies the given angle.
The tangent PT and the radius OT meet at the point of contact T at \(90^\circ\), so angle OTP \(= 90^\circ\) for every tangent. The tempting angle OPT is at the outside point P, where the angle depends on how far P is from O. Angle TOP is at the centre, and the right angle at T needs no condition about diameters.
Angle ACB \(= 90^\circ\) (angle in a semicircle), so angle ABC \(= 180^\circ - 90^\circ - 38^\circ = 52^\circ\). Angle ABD \(= 90^\circ\) (tangent meets radius), so \(x = 90^\circ - 52^\circ = 38^\circ\). The tempting \(52^\circ\) is angle ABC, or angle ADB, not the angle between BC and the tangent. \(90^\circ\) is the whole angle ABD, and \(19^\circ\) halves \(38^\circ\).
OP is the hypotenuse: \(OT = \sqrt{10^2 - 8^2} = \sqrt{36} = 6\) cm, so the diameter is \(2 \times 6 = 12\) cm. The tempting 6 cm stops at the radius. 12.8 cm adds the squares (\(\sqrt{164}\)), treating OP as a shorter side, and 25.6 cm doubles that wrong value.
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.