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GCSE Maths Circle Theorems (Higher): check-up and exam-style questions

Part of the GCSE Maths circle theorems chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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Exercise 26Z Check-up (non-calculator)

Non-calculator

AB is a diameter of the circle. C is a point on the circle. Angle CAB = 42°. Work out the size of angle ABC. Give a reason for each step.

48° (angle in a semicircle is 90°; the angles in a triangle add up to 180°)

Key words. A diameter is a chord through the centre. An angle in a semicircle is the angle made at the circle by lines drawn to both ends of a diameter.

Method. A diameter makes a right angle with every point on the circle. Use that right angle, then the angle sum of a triangle.

Mark the triangle ABC. The diameter AB and the point C on the circle make a triangle. We are looking for angle ABC in it.
Angle CAB = 42°, angle ABC = ?
Find angle ACB. The line AB is a diameter, so the angle at C on the circle is an angle in a semicircle.
Angle ACB = 90° (reason: angle in a semicircle is 90°)
Use the angle sum of the triangle. The three angles of triangle ABC add up to 180°, so subtract the two we know.
Angle ABC = 180° − 90° − 42°
Angle ABC = 48° (reason: the angles in a triangle add up to 180°)

Check. 42° + 48° + 90° = 180°. ✓

Watch out. The right angle is at the point on the circle (C), not at the end of the diameter.

Done

A, B and C are points on a circle with centre O. Angle ACB = 48°. Work out the size of angle AOB. Give a reason for your answer.

96° (the angle at the centre is twice the angle at the circumference)

Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.

Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.

Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.
Centre: angle AOB = ?
Circumference: angle ACB = 48°
Choose the rule. We are going from the circumference to the centre, so we use the circle theorem for the angle at the centre.
angle at the centre = 2 × angle at the circumference
Double it. The angle at the centre is twice the angle at the circumference.
Angle AOB = 2 × 48° = 96°

Check. Half of 96° is 48°. ✓

Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.

Done

A, B and C are points on a circle with centre O. Angle ACB = 38°. Work out the size of angle OAB. Give a reason for each step.

52° (the angle at the centre is twice the angle at the circumference; the base angles of an isosceles triangle are equal)

Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.

Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.

Find angle AOB. Angle at the centre and angle at the circumference are both on the same arc AB.
Angle AOB = 2 × 38° = 76° (reason: the angle at the centre is twice the angle at the circumference)
Spot the isosceles triangle. OA and OB are radii, so they are equal. Triangle OAB is isosceles and its base angles are equal.
Angle OAB = angle OBA (the base angles of an isosceles triangle are equal)
Use the angle sum in triangle OAB. The angles add up to 180°. Take off the angle at O and share the rest equally.
2 × angle OAB = 180° − 76° = 104°
Angle OAB = 104° ÷ 2 = 52°

Check. 52° + 52° + 76° = 180°. ✓ Also 52 = 90 − 38. ✓

Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.

Done

A, B, C and D are points on a circle. Angle ACB = 45°. Work out the size of angle ADB. Give a reason for your answer.

45° (angles in the same segment are equal)

Key words. A chord cuts a circle into two segments. Two angles are in the same segment if their corners are on the same side of the chord.

Method. Two angles made from the same chord, with their corners on the same side of it, are in the same segment and are equal.

Find the chord that both angles are drawn from. Both angles have their arms ending at A and B. So both are made from the chord AB.
Angle ACB and angle ADB both stand on chord AB.
Check the two points are on the same side of the chord. C and D are both on the same side of the line AB, so they lie in the same segment.
C and D are in the same segment of the circle.
Use the theorem. Angles in the same segment are equal.
Angle ADB = angle ACB = 45°

Check. Angle at the centre on the same arc would be 2 × 45° = 90° for either point. ✓

Watch out. Make sure both corners are on the same side of the chord. Angles on opposite sides of the chord add up to 180°, not equal.

Done

ABCD is a cyclic quadrilateral. Angle DAB = 95°. Work out the size of angle BCD. Give a reason for your answer.

85° (opposite angles of a cyclic quadrilateral add up to 180°)

Key words. A cyclic quadrilateral is a four-sided shape with all four corners on a circle. Opposite angles are at corners that are not next to each other.

Method. In a cyclic quadrilateral, opposite angles add up to 180°. Subtract the known angle from 180°.

Check the angles are opposite. In a quadrilateral ABCD, A is opposite C and B is opposite D. Opposite corners are not next to each other.
Angle DAB and angle BCD are opposite corners.
Write the cyclic quadrilateral fact. All four corners are on the circle, so opposite angles add up to 180°.
Angle DAB + angle BCD = 180°
Subtract. Take the known angle away from 180°.
Angle BCD = 180° − 95° = 85°

Check. 95° + 85° = 180°. ✓

Watch out. Opposite means the corner across the quadrilateral. Neighbouring angles do not add up to 180°.

Done

TA and TB are tangents to a circle with centre O. Angle ATB = 70°. Work out the size of angle AOB. Give a reason for each step.

110° (a tangent is perpendicular to the radius at the point of contact; the angles in a quadrilateral add up to 360°)

Method. The two radii and the two tangents make a quadrilateral OATB with two right angles. Use the angle sum 360°.

Draw the radii OA and OB. They go to the two points of contact and make a quadrilateral OATB with the two tangents.
quadrilateral OATB
Mark the right angles. Each tangent is perpendicular to the radius drawn to its point of contact.
Angle OAT = 90° and angle OBT = 90° (a tangent is perpendicular to the radius at the point of contact)
Use the quadrilateral OATB. Its four angles add up to 360°. Subtract the three we know.
Angle AOB = 360° − 90° − 90° − 70° = 110° (the angles in a quadrilateral add up to 360°)

Check. 360° = 360°. ✓ A shortcut for later: angle AOB + angle ATB = 180°.

Watch out. The quadrilateral OATB is a kite. Its two right angles are at A and B, opposite each other.

Done

TA is a tangent to a circle with centre O, touching it at A. The radius of the circle is 8 cm and OT = 17 cm. Work out the length of TA.

15 cm

Method. The radius meets the tangent at a right angle, so triangle OAT is right-angled and we can use Pythagoras' theorem.

Find the right angle. A tangent is perpendicular to the radius at the point of contact.
Angle OAT = 90° (a tangent is perpendicular to the radius at the point of contact)
Write Pythagoras for triangle OAT. The longest side is OT, opposite the right angle.
OT² = OA² + TA²
17² = 8² + TA²
Rearrange. Take the known square away from both sides.
TA² = 289 − 64 = 225
Take the square root. A length is positive.
TA = √225 = 15 cm

Check. 8² + 15² = 289, which is 17². ✓

Watch out. OT is the hypotenuse, so it is the one you square and then subtract from, not add to.

Done

PQ is a tangent to the circle at T. A and B are points on the circle. Angle TBA = 44°. Work out the size of angle QTA. Give a reason for your answer.

44° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Key words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.

Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).

Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.
Angle TBA is in the segment on the other side of TA from angle QTA.
Find the matching angle. The alternate segment is on the other side of the chord TA from angle QTA. That is where B is, so angle TBA is the matching angle.
angle TBA is in the alternate segment
Use the alternate segment theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.
Angle QTA = angle TBA = 44° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. Angle at the centre on arc TA would be 2 × 44° = 88°. Both angles are half of it. ✓

Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.

Done

AB is a chord of a circle with centre O and radius 10 cm. AB = 16 cm. M is the point where the perpendicular from O meets the chord. Work out the distance OM.

6 cm

Key words. To bisect means to cut exactly in half. A perpendicular meets another line at 90°.

Method. The perpendicular from the centre cuts the chord in half, which makes a right-angled triangle with the radius as hypotenuse.

Find the length AM. The perpendicular from the centre to a chord bisects (cuts in half) the chord.
AM = 16 ÷ 2 = 8 cm (the perpendicular from the centre to a chord bisects the chord)
Write Pythagoras for triangle OMA. The angle at M is 90°. OA is a radius, and it is the longest side (the hypotenuse).
OA² = OM² + AM²
10² = OM² + 8²
Rearrange. Take the known square from both sides.
OM² = 100 − 64 = 36
Square root.
OM = √36 = 6 cm

Check. 6² + 8² = 36 + 64 = 100 = 10². ✓

Watch out. Use half the chord in the triangle, not the whole chord.

Done

TA and TB are tangents to a circle with centre O. C is a point on the circle. Angle ATB = 70°. Work out the size of angle ACB. Give a reason for each step.

55° (a tangent is perpendicular to the radius at the point of contact; the angles in a quadrilateral add up to 360°; the angle at the centre is twice the angle at the circumference)

Method. Use the tangent-radius right angles to find the angle at the centre, then halve it for the angle at the circumference.

Find the angles at A and B. Each tangent is perpendicular to the radius at its point of contact.
Angle OAT = angle OBT = 90° (a tangent is perpendicular to the radius at the point of contact)
Find the angle at the centre. Quadrilateral OATB has angles that add up to 360°.
Angle AOB = 360° − 90° − 90° − 70° = 110° (the angles in a quadrilateral add up to 360°)
Halve it. Angle ACB is at the circumference and stands on the same arc AB as angle AOB.
Angle ACB = 110° ÷ 2 = 55° (the angle at the centre is twice the angle at the circumference)

Check. Second route (alternate segment): triangle TAB is isosceles, so angle TAB = (180° − 70°) ÷ 2 = 55°, and by the alternate segment theorem this equals angle ACB. ✓

Watch out. There are two circle theorems used here. Name each one as you use it.

Done

Total 59 marks. Allow about 59 minutes.

Exercise 26X Exam-style questions

Non-calculator

FA, B and C are points on a circle with centre O. Angle AOB = 104°. Work out the size of angle ACB. Give a reason for your answer.[2 marks]

52° (the angle at the centre is twice the angle at the circumference)

Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.

Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.

Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.
Centre: angle AOB = 104°
Circumference: angle ACB = ?
Choose the rule. We are going from the centre to the circumference, so we use the circle theorem the other way round.
angle at the circumference = angle at the centre ÷ 2
Halve it. The angle at the centre is twice the angle at the circumference, so the angle at the circumference is half the angle at the centre.
Angle ACB = 104° ÷ 2 = 52°

Check. Double 52° is 104°. ✓

Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.

Done

FAB is a diameter of the circle. Angle CAB = (2x + 14)° and angle ABC = (5x − 8)°. Work out the value of x.[4 marks]

x = 12

Method. The angle in a semicircle is 90°, so the other two angles of the triangle add up to 90°. That gives an equation in x.

Find angle ACB. The line AB is a diameter, so angle ACB is an angle in a semicircle.
Angle ACB = 90° (angle in a semicircle is 90°)
Write an equation. The three angles of triangle ABC add up to 180°. With the right angle taken away, the other two must add to 90°.
(2x + 14) + (5x − 8) = 90
Collect the like terms. Add the x terms together and the numbers together.
7x + 6 = 90
Subtract 6 from both sides. We want the number terms on the other side.
7x = 90 − 6
7x = 84
Divide both sides by 7. 7x means 7 × x, so dividing by 7 leaves x on its own.
x = 84 ÷ 7 = 12
Check the answer by working out both angles. Put x back into each angle. They should add up to 90°.
2 × 12 + 14 = 38° and 5 × 12 + -8 = 52°
38° + 52° = 90° ✓

Check. The right angle plus 38° plus 52° is 90° + 90° = 180°. ✓

Watch out. The equation is "angles add to 90°", not 180°, because the right angle is already used up.

Done

FABCD is a cyclic quadrilateral. Angle DAB = 100°. Work out the size of angle BCD. Give a reason for your answer.[2 marks]

80° (opposite angles of a cyclic quadrilateral add up to 180°)

Key words. A cyclic quadrilateral is a four-sided shape with all four corners on a circle. Opposite angles are at corners that are not next to each other.

Method. In a cyclic quadrilateral, opposite angles add up to 180°. Subtract the known angle from 180°.

Check the angles are opposite. In a quadrilateral ABCD, A is opposite C and B is opposite D. Opposite corners are not next to each other.
Angle DAB and angle BCD are opposite corners.
Write the cyclic quadrilateral fact. All four corners are on the circle, so opposite angles add up to 180°.
Angle DAB + angle BCD = 180°
Subtract. Take the known angle away from 180°.
Angle BCD = 180° − 100° = 80°

Check. 100° + 80° = 180°. ✓

Watch out. Opposite means the corner across the quadrilateral. Neighbouring angles do not add up to 180°.

Done

FABCD is a cyclic quadrilateral. Angle ABC = (4x + 30)° and angle CDA = (2x + 60)°.[4 marks]

(a)Work out the value of x.

x = 15

Method. Opposite angles of a cyclic quadrilateral add up to 180°. Write that as an equation, solve for x, then find the angles.

Write the angle fact. B and D are opposite corners of a cyclic quadrilateral, so the angles add up to 180°.
(4x + 30) + (2x + 60) = 180 (opposite angles of a cyclic quadrilateral add up to 180°)
Collect like terms. Add the x terms together, then the numbers.
6x + 90 = 180
Subtract 90 from both sides. We want the number terms on the other side.
6x = 180 − 90
6x = 90
Divide both sides by 6. 6x means 6 × x, so dividing by 6 leaves x on its own.
x = 90 ÷ 6 = 15
Work out the two angles. Put x back into each expression.
Angle ABC = 4 × 15 + 30 = 90°
Angle CDA = 2 × 15 + 60 = 90°

Check. 90° + 90° = 180°. ✓

Watch out. Put each expression in brackets and add them. Do not try to make one expression equal the other.

(b)Work out angles ABC and CDA.

90° and 90°

Method. Opposite angles of a cyclic quadrilateral add up to 180°. Write that as an equation, solve for x, then find the angles.

Write the angle fact. B and D are opposite corners of a cyclic quadrilateral, so the angles add up to 180°.
(4x + 30) + (2x + 60) = 180 (opposite angles of a cyclic quadrilateral add up to 180°)
Collect like terms. Add the x terms together, then the numbers.
6x + 90 = 180
Subtract 90 from both sides. We want the number terms on the other side.
6x = 180 − 90
6x = 90
Divide both sides by 6. 6x means 6 × x, so dividing by 6 leaves x on its own.
x = 90 ÷ 6 = 15
Work out the two angles. Put x back into each expression.
Angle ABC = 4 × 15 + 30 = 90°
Angle CDA = 2 × 15 + 60 = 90°

Check. 90° + 90° = 180°. ✓

Watch out. Put each expression in brackets and add them. Do not try to make one expression equal the other.

Done

FA, B, C and D are points on a circle. The chords AC and BD meet at E. Angle BAC = 40° and angle AEB = 95°. Work out the size of angle ACD. Give a reason for each step.[4 marks]

45° (the angles in a triangle add up to 180°; angles in the same segment are equal)

Method. Use the triangle angle sum first to find an angle that is in the same segment as the one we want.

Pick out triangle ABE. The chords cross at E. Triangle ABE contains the angle BAC and the angle AEB that we are given.
triangle ABE: angle BAE = 40°, angle AEB = 95°
Find angle ABE (that is, angle ABD). Triangle ABE has angle sum 180°. We know two of its angles.
Angle ABD = 180° − 40° − 95° = 45° (the angles in a triangle add up to 180°)
Match it with an angle in the same segment. Angles ABD and ACD are both made from the chord AD, with B and C on the same side of it.
Angle ACD = angle ABD = 45° (angles in the same segment are equal)

Check. In triangle DEC the angles are 40° (angle BDC, same segment as BAC), 45° and 95° (vertically opposite AEB): 40 + 45 + 95 = 180. ✓

Watch out. A "same segment" pair needs the same chord. Name the chord before you say the angles are equal.

Done

FTA and TB are tangents to a circle with centre O. Angle ATB = 66°. Work out the size of angle OAB. Give a reason for each step.[4 marks]

33° (a tangent is perpendicular to the radius at the point of contact; the angles in a quadrilateral add up to 360°; the base angles of an isosceles triangle are equal)

Method. First find angle AOB from the kite OATB, then use the isosceles triangle made by two radii.

Find angle AOB. Quadrilateral OATB has right angles at A and B (tangent perpendicular to radius). Its angles add up to 360°.
Angle AOB = 360° − 90° − 90° − 66° = 114° (a tangent is perpendicular to the radius at the point of contact; the angles in a quadrilateral add up to 360°)
Spot the isosceles triangle. OA and OB are radii, so triangle OAB is isosceles and its base angles are equal.
Angle OAB = angle OBA (the base angles of an isosceles triangle are equal)
Use the angle sum in triangle OAB. Subtract the angle at O and share what is left.
Angle OAB = (180° − 114°) ÷ 2 = 66° ÷ 2 = 33° (the angles in a triangle add up to 180°)

Check. Angle OAT = 90° and angle OAB + angle BAT should be 90°. Triangle TAB is isosceles with base angles (180° − 66°) ÷ 2 = 57°, and 33° + 57° = 90°. ✓

Watch out. Both triangles OAB and TAB are isosceles. Be careful about which one you are using.

Done

FTA and TB are tangents to a circle with centre O. TA = (3x + 5) cm and TB = (4x − 1) cm.[4 marks]

(a)Work out the value of x.

x = 6

Method. Tangents from an external point are equal in length, so the two expressions are equal. Solve the equation, then substitute.

Use the tangent fact. Tangents drawn from the same point to a circle are equal in length.
TA = TB (tangents from an external point are equal in length)
Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.
4x − 1 = 3x + 5 (written the other way round so that the x terms stay positive)
Subtract 3x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.
4x − 1 − 3x = 3x + 5 − 3x
x − 1 = 5
Add 1 to both sides. We want the number terms on the other side.
x = 5 + 1
x = 6
Read off x.
x = 6
Work out the length. Put x back into the expression for TA.
TA = 3 × 6 + 5 = 23 cm

Check. TB = 4 × 6 − 1 = 23 cm, the same as TA. ✓

Watch out. Check the final length is positive. A negative length means a mistake.

(b)Work out the length of TA.

23 cm

Method. Tangents from an external point are equal in length, so the two expressions are equal. Solve the equation, then substitute.

Use the tangent fact. Tangents drawn from the same point to a circle are equal in length.
TA = TB (tangents from an external point are equal in length)
Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.
4x − 1 = 3x + 5 (written the other way round so that the x terms stay positive)
Subtract 3x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.
4x − 1 − 3x = 3x + 5 − 3x
x − 1 = 5
Add 1 to both sides. We want the number terms on the other side.
x = 5 + 1
x = 6
Read off x.
x = 6
Work out the length. Put x back into the expression for TA.
TA = 3 × 6 + 5 = 23 cm

Check. TB = 4 × 6 − 1 = 23 cm, the same as TA. ✓

Watch out. Check the final length is positive. A negative length means a mistake.

Done

FPQ is a tangent to the circle at T. A and B are points on the circle. Angle QTA = 35° and angle PTB = 75°. Give a reason for each answer.[5 marks]

(a)Work out angle TBA

35° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Method. Use the alternate segment theorem: the angle between the tangent and a chord equals the angle in the alternate segment.

Look at the tangent angle on the chord TA. Angle QTA is between the tangent and chord TA. The alternate segment is on the other side of TA, and it contains B.
angle QTA = 35°, and the matching angle is angle TBA
Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.
Angle TBA = angle QTA = 35° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. The matching angle at the centre on arc TA would be 70°. ✓

Watch out. The alternate segment is the one across the chord from the tangent angle.

(b)Work out angle TAB

75° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Method. Use the alternate segment theorem again, this time with the chord TB.

Look at the tangent angle on the chord TB. Angle PTB is between the tangent and chord TB. The alternate segment is on the other side of TB, and it contains A.
angle PTB = 75°, and the matching angle is angle TAB
Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.
Angle TAB = angle PTB = 75° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. The matching angle at the centre on arc TB would be 150°. ✓

Watch out. Use the angle on the opposite side of chord TB from PTB, which is at A.

(c)Work out angle ATB

70° (the angles in a triangle add up to 180°)

Method. The three angles of triangle ATB add up to 180°.

Write the two angles we know. We found them in parts (a) and (b).
Angle TBA = 35° and angle TAB = 75°
Subtract from 180°. The angles in a triangle add up to 180°.
Angle ATB = 180° − 35° − 75° = 70°

Check. Also angle PTB + angle ATB + angle QTA = 75 + 70 + 35 = 180 (angles on a straight line). ✓

Watch out. You can also find this angle from the straight line PQ.

Done

FPQ is a tangent at T to a circle with centre O. Angle QTA = 32°. Work out the size of angle TOA. Give a reason for each step.[4 marks]

64° (the angle between a tangent and a chord is equal to the angle in the alternate segment; the angle at the centre is twice the angle at the circumference)

Method. First use the alternate segment theorem to get an angle at the circumference, then double it for the angle at the centre.

Find the angle in the alternate segment. Angle QTA is between the tangent and the chord TA.
Angle TBA = 32° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Name the arc. Angle TOA is at the centre and angle TBA is at the circumference. Both stand on the arc TA.
arc TA
Use the angle at the centre. The angle at the centre is twice the angle at the circumference.
Angle TOA = 2 × 32° = 64° (the angle at the centre is twice the angle at the circumference)

Check. Second route: angle OTQ = 90° (tangent and radius), so angle OTA = 90° − 32° = 58°. Triangle OTA is isosceles, so angle OAT = 58° and angle TOA = 180° − 2 × 58° = 64°. ✓

Watch out. You can also find this answer without the alternate segment theorem, using the right angle at T (see the check).

Done

FAB is a chord of a circle with centre O and radius 13 cm. AB = 24 cm. M is the point where the perpendicular from O meets the chord. Work out the distance OM.[3 marks]

5 cm

Key words. To bisect means to cut exactly in half. A perpendicular meets another line at 90°.

Method. The perpendicular from the centre cuts the chord in half, which makes a right-angled triangle with the radius as hypotenuse.

Find the length AM. The perpendicular from the centre to a chord bisects (cuts in half) the chord.
AM = 24 ÷ 2 = 12 cm (the perpendicular from the centre to a chord bisects the chord)
Write Pythagoras for triangle OMA. The angle at M is 90°. OA is a radius, and it is the longest side (the hypotenuse).
OA² = OM² + AM²
13² = OM² + 12²
Rearrange. Take the known square from both sides.
OM² = 169 − 144 = 25
Square root.
OM = √25 = 5 cm

Check. 5² + 12² = 25 + 144 = 169 = 13². ✓

Watch out. Use half the chord in the triangle, not the whole chord.

Done

FA circular pipe has radius 10 cm. Water fills the pipe to a level so that the water surface is a chord of length 16 cm. Work out the depth of the water at its deepest point (the vertical distance from the chord to the lowest part of the circle).[4 marks]

4 cm

Method. Use the perpendicular from the centre to the chord, then take the distance from the radius.

Draw the picture. The surface of the water is a chord. The lowest point of the water is at the bottom of the circle, directly below the centre.
Chord = 16 cm, radius = 10 cm
Find the distance from the centre to the chord. The perpendicular from the centre bisects the chord, which gives a right-angled triangle with half the chord and the radius.
Half the chord = 8 cm
Distance² = 10² − 8² = 100 − 64 = 36
Distance = 6 cm
Work out the depth. The depth is the radius take away the distance from the centre to the chord.
Depth = 10 − 6 = 4 cm

Check. Water is 4 cm deep and the centre is 6 cm above the surface: 4 + 6 = 10 cm, the radius. ✓

Watch out. The depth is not the distance from the centre to the chord. It is the radius minus that distance.

Done

FTU is a tangent to a circle with centre O, touching the circle at A. B is a point on the circle. Angle UAB = 40°. Give a reason for each step.[5 marks]

(a)Work out angle OAB

50° (a tangent is perpendicular to the radius at the point of contact)

Method. A tangent is perpendicular to the radius at the point of contact, so angle OAU is a right angle.

Find the right angle. Radius OA meets the tangent TU at the point of contact A.
Angle OAU = 90° (a tangent is perpendicular to the radius at the point of contact)
Subtract. Angle OAU is made of angle OAB and angle BAU.
Angle OAB = 90° − 40° = 50°

Check. 50° + 40° = 90°. ✓

Watch out. The 90° is between the tangent and the radius, so it contains the angle OAB plus the angle UAB.

(b)Work out angle AOB

80° (the base angles of an isosceles triangle are equal; the angles in a triangle add up to 180°)

Method. Triangle OAB has two equal sides (both radii), so it is isosceles.

Write the base angles. OA and OB are radii, so the base angles are equal.
Angle OBA = angle OAB = 50° (the base angles of an isosceles triangle are equal)
Use the angle sum. Subtract the two base angles from 180°.
Angle AOB = 180° − 50° − 50° = 80° (the angles in a triangle add up to 180°)

Check. Alternate segment theorem: the angle in the alternate segment is 40°, and the angle at the centre is 2 × 40° = 80°. ✓

Watch out. The apex angle of the isosceles triangle is at the centre O.

Done

FAB is a diameter of a circle with centre O. TU is a tangent to the circle at A. C is a point on the circle. Angle UAC = 36°. Give a reason for each answer.[6 marks]

(a)Work out angle CAB.

54° (a tangent is perpendicular to the radius at the point of contact)

Method. A tangent is perpendicular to the radius at the point of contact. AB is a diameter, so it contains the radius OA.

Find angle UAB. The diameter AB is along the radius OA, and the tangent TU touches the circle at A.
Angle UAB = 90° (a tangent is perpendicular to the radius at the point of contact)
Subtract. Angle UAB is made of angle UAC and angle CAB side by side.
Angle CAB = 90° − 36° = 54°

Check. 36° + 54° = 90°. ✓

Watch out. The right angle is between the tangent and the diameter, so it contains both angles.

(b)Work out angle ACB.

90° (angle in a semicircle is 90°)

Method. AB is a diameter, so angle ACB is an angle in a semicircle.

Spot the diameter. The two arms CA and CB go to the two ends of the diameter AB.
Angle ACB = 90° (angle in a semicircle is 90°)
Check the triangle. Triangle ABC now has a right angle at C.
angle ACB = 90°

Check. Angle CAB + angle ACB = 54° + 90° = 144°, which is less than 180°, so there is room for angle ABC. ✓

Watch out. The right angle is at C, on the circle.

(c)Work out angle ABC.

36° (the angles in a triangle add up to 180°; the angle between a tangent and a chord is equal to the angle in the alternate segment)

Method. Use the angle sum in triangle ABC, then check with the alternate segment theorem.

Use the angles we know. Angle CAB is from part (a) and angle ACB is from part (b).
Angle ABC = 180° − 54° − 90° = 36° (the angles in a triangle add up to 180°)
Check with the alternate segment theorem. Angle UAC is between the tangent and the chord AC. The angle in the alternate segment on the other side of AC is angle ABC.
Angle ABC = angle UAC = 36° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. Both methods give 36°. ✓

Watch out. When two different routes give the same answer, you can be confident it is right.

Done

RA, B, C and D are points on a circle. The chords AC and BD cross at E. Prove that triangles ABE and DCE are similar.[4 marks]

Angle AEB = angle DEC (vertically opposite) and angle BAE = angle CDE (angles in the same segment), so the triangles are similar (two equal angles).

Method. Show that two angles of triangle ABE equal two angles of triangle DCE. Then the triangles are similar.

Find a pair of angles at E. The chords cross at E, so angle AEB and angle DEC are opposite each other.
angle AEB = angle DEC (vertically opposite angles are equal)
Find a second pair of angles. Angle BAE is the same as angle BAC, and angle CDE is the same as angle CDB. Both are made from the chord BC with A and D on the same side of it.
angle BAE = angle CDE (angles in the same segment are equal)
Conclude. Two pairs of equal angles are enough to show triangles are similar (the third pair must then be equal too).
So triangle ABE is similar to triangle DCE. ✓

Check. Test: arcs AB = 100°, BC = 70°, CD = 110°, DA = 80° give angle BAC = angle BDC = 35° and angle AEB = 105° = angle DEC. ✓

Watch out. Two equal angles (AA) are enough for similar triangles. Give the reason for each pair.

Done

RTA and TB are tangents to a circle with centre O. C is a point on the major arc AB. Prove that angle TAB = angle ACB.[4 marks]

Triangle TAB is isosceles (tangents from T are equal), and angle TAB = angle ACB by the alternate segment theorem. (Or both equal 90° − ½ angle ATB.)

Method. Use the alternate segment theorem directly, or check with the kite and the centre theorem.

Spot the tangent and the chord. TA is a tangent that touches the circle at A, and AB is a chord from the point of contact.
angle TAB is between the tangent TA and the chord AB
Find the alternate segment. It is on the other side of the chord AB from the tangent angle. C is on the major arc, which is on that side.
angle ACB is in the alternate segment
Use the theorem.
angle TAB = angle ACB (the angle between a tangent and a chord is equal to the angle in the alternate segment). ✓
Second route (check). Let angle ATB = t. Triangle TAB is isosceles, so angle TAB = (180° − t) ÷ 2. The kite OATB gives angle AOB = 180° − t, so angle ACB = (180° − t) ÷ 2.
Both equal 90° − ½t.

Check. Test: t = 64° gives 58° both ways. ✓

Watch out. Name the tangent, the chord and the alternate segment.

Done

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