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In order to access this I need to be confident with:

Parts of a circle Angles in polygons Angles on a straight line Angles around a point Angles in parallel lines Types of trianglesThis topic is relevant for:

Here we will learn about **angles in a semicircle***,* including their application, proof, and using them to solve more difficult problems.

There are also circle theorem worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

**Angles in a semicircle are 90 degrees**:

Angles in a semicircle are created when you join the two ends of the diameter to one point on the arc using chords.

In the diagram above, AB is the diameter of a circle that divides the circle into two. The two points A and B are joined to another point C on the circumference using two chords. This triangle is a right-angle triangle with the 90 degree angle touching the arc. This is the same for any point that is placed on the arc.

If the point is placed on another point in a semicircle, the result would be the same (the two chords would meet at the circumference at \pmb{90} degrees). **The angle in a semicircle is 90 degrees**.

- The
**diameter**of the circle is the width of a circle, through the centre. The diameter is twice the length of the**radius**of a circle. In a semicircle, the diameter is the straight edge. - A
**chord**is a straight line that meets the circumference in two places. The longest chord in a circle is the**diameter**. - A
**semicircle**is half of a circle.

**Subtended angles (in a semicircle)**:

An angle within a semicircle is created by two chords meeting at the two ends of the diameter. The diagrams below show the angle subtended by diameter AC from point B for two different semicircles:

To be able to prove this theorem, you do not need to know any other circle theorem. You just need to be confident with angles in a triangle.

Step | Diagram | Description |

1 | We start with the diagram of a circle with triangle ABC where A, B, and C are points on the circumference and AB is the diameter of the circle, going through the centre. | |

2 | Connect the centre to the point C to create two smaller triangles. | |

3 | Let us inspect triangle OAC. | |

4 | OA and OC are radii and so triangle OAC is isosceles. This means that the two angles at A and C are equal, so we have labelled them both x. | |

5 | Let us now inspect triangle OBC. | |

6 | OB and OC are radii and so triangle OBC is also isosceles. This means that the two angles at B and C are equal, so we have labelled them both y. We cannot label them both x as they may not be the same as in triangle OAB. | |

7 | So now we have the triangle with the following angles. | |

8 | We can simplify the angle at C (angle ACB) to equal x + y as this angle is the sum of the two angles x and y from the original two triangles OAB and OBC. | |

9 | For the final step in this proof, we need to form and solve an equation to show that x + y = 90°. | As the sum of angles in a triangle is equal to 180°, we can state: As x + y = 90°, we can now state that the angle in a semicircle is 90 degrees. |

In order to use the fact that angles in a semicircle equal 90° :

**Locate the key parts of the circle for the theorem.****Use other angle facts to determine angles within the triangle.****Use the angles in a semicircle theorem to state the other missing angle.**

Get your free angle in a semicircle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free angle in a semicircle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREETangent of a circle is one of 7 circle theorems you will need to know. You may find it helpful to start with our main circle theorems page and then look in detail at the rest.

ABCD is an arrowhead where C is the centre of the circle and A, B, and D lie on the circumference. Calculate the size of angle BAD :

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle ABC = 67°
- AB is a diameter
- AC and BC are chords
- The angle CAB = θ

2**Use other angle facts to determine angles within the triangle**.

We already know that ABC=67° so we do not need to use any other angle fact to determine this angle for this example.

3**Use the angles in a semicircle theorem to state the other missing angle**.

As the angle in a semicircle is equal to 90° , angle ACB = 90° . We can therefore use the fact that angles in a triangle total 180° to calculate the size of angle BAC :

BAC = 180 - (90 + 67)

BAC=23°

ABCD is a cyclic quadrilateral. ACD is an isosceles triangle. Calculate the size of angle ACB .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle BAD = 107°
- AC is a diameter
- AB and BC are chords
- The angle ACB = θ

**Use other angle facts to determine angles within the triangle**.

Triangle ACD is interesting because it is an isosceles triangle but also a right angle triangle with the hypotenuse AC .

Using the theorem that the angle in a semicircle is 90° , we can say that angle ADC is equal to 90° .

As the triangle is isosceles, the other two angles are equal to (180 - 90) ÷ 2 = 45° . This is important because we now know that angle DAC is equal to 45° , and so we can calculate the size of angle BAC which is in the triangle containing θ.

Angle BAC is therefore equal to:

BAC = 107 - 45

BAC = 62°

We now have the following information added to the diagram:

**Use the angles in a semicircle theorem to state the other missing angle**.

As the angle in a semicircle is equal to 90° , angle ABC = 90° . We can therefore use the fact that **angles in a triangle total 180° ** to calculate the size of the inscribed angle BCA :

BCA = 180 - (90 + 62)

BAC = 28°

Points A, B, C, and D lie on the circumference of the circle. AD is a diameter of the circle. BD and AC intersect at the point E . Calculate the size of the angle CDB .

**Locate the key parts of the circle for the theorem**.

Here we have:

- AD is a diameter
- AB, BD, AC and CD are chords
- The angle BEA = 48°
- The angle CAD = 36°
- The angle BDC = θ

**Use other angle facts to determine angles within the triangle**.

In order to find the angle θ, we need to know the size of the angle EDA . BED is a straight line and so, as **angles on a straight line total 180° **, we can calculate angle AED :

AED = 180 - 48

AED = 132°

We can now calculate angle ADE as **angles in a triangle total 180° **:

ADE = 180 - (132 + 36)

ADE = 12°

**Use the angles in a semicircle theorem to state the other missing angle**.

As the angle in a semicircle is equal to 90° , angle ACD = 90° , we can, therefore, use the fact that** angles in a triangle total 180° **to calculate the size of angle CDA , and hence angle CDE :

CDA = 180 - (90 + 36)

CDA = 54°

CDE = 54 - 12

CDE = 42°

AC and BD are diameters of the circle with centre O . BC and AD are parallel. Calculate the size of the angle OAB .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle COD = 58°
- BD and AC are diameters
- OA, OB, OC, and OD are radii
- The angle BAC = θ

**Use other angle facts to determine angles within the triangle**.

AOD is an isosceles triangle as OA and OD are radii. This means that the other two angles are the same and so:

OAD = (180 - 122) ÷ 2

OAD = 29°

**Use the angles in a semicircle theorem to state the other missing angle**.

As the angle in a semicircle is equal to 90° , angle BAD = 90° . Angle OAB is therefore equal to:

OAB = 90 - 29

OAB = 61°

ABCD are points on a circle with centre O . EF is a tangent to the circle through the point C . BD and EF are parallel. Angle DCE = 63° .

Calculate the size of angle BAC .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle DCE = 63°
- AC is a diameter
- The angle BAC = θ

**Use other angle facts to determine angles within the triangle**.

The alternate segment theorem states that angles in the alternate segment are equal. The angle CBD is, therefore, equal to 63° . Also, as the lines BD and EF are parallel, the line AC is perpendicular to the line BD . So the angles at their intersection G are equal to 90° .

BCG is a triangle. As we know two of the three angles, we can now calculate the size of angle BCG , which will help us determine angle CAB as the angles are congruent.

BCG = 180 - (90 + 63)

BCG = 27°

**Use the angles in a semicircle theorem to state the other missing angle**.

As the angle in a semicircle is equal to 90° , angle ABC = 90° . Angle BAC is therefore equal to:

BAC = 180 - (90 + 27)

BAC = 63°

A, B, C, D and E are points on the circle with centre O . Points A and O lie on a line of symmetry. Calculate the size of the angle DAE .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle DOC = 90°
- BE is a diameter
- The angle DAE = θ

**Use other angle facts to determine angles within the triangle**.

The **angle at the centre is twice the angle at the circumference** and so angle DAC = 45° .

As the shape is symmetrical, the angle OAD is half the angle DAC . So OAD = 22.5° .

**Use the angles in a semicircle theorem to state the other missing angle**.

As the angle in a semicircle is 90° , the angle BAE = 90° . Due to the shape being symmetrical, angle OAE is half of the angle BAE . Angle OAE = 45° .

We can now calculate the angle DAE :

DAE = 45 - 22.5

DAE = 22.5°

**Cyclic quadrilateral**

The angles that are either end of the diameter total 180° as if the triangle were a cyclic quadrilateral. They should total 90° as the angle in a semicircle is 90° .

**Isosceles triangles**

The triangle is assumed to be isosceles and so the angles at either end of the diameter are the same.

**A diameter or a chord?**

The angle at the circumference is assumed to be 90° when the associated chord does not intersect the centre of the circle and so the diagram does not show a semicircle.

**Incorrect angle**

The angle that is established as 90° , is labelled as an angle by the diameter, and not the subtended angle at the circumference.

1. The points A, B, and C are on the circumference of a circle. Calculate the size of angle BCA .

127^{\circ}

37^{\circ}

53^{\circ}

90^{\circ}

- ABC = 90^{\circ} (angles in a semicircle)
- BCA = 180 – (90+53) = 37^{\circ}

2. The points A, B, and C are on the circumference of a circle. Angle ABC = 79^{\circ}. Calculate the size of angle BCA .

56^{\circ}

101^{\circ}

79^{\circ}

50.5^{\circ}

- BAC = BCD = 90^{\circ} (angles in a semicircle)
- ABC = (180 – 90)\div2 = 45^{\circ} (isosceles triangle)
- CBD = 79 – 45 = 34^{\circ}
- CDB = 180 – (90+34) = 56^{\circ}

3. The points A, B, C, and D are on the circumference of a circle. AC and BD are chords that intersect at the point E . Calculate the size of angle DAE .

63^{\circ}

35^{\circ}

55^{\circ}

27^{\circ}

- AED = 35^{\circ} (vertically opposite angles are equal)
- ADE = 90^{\circ} (angles in a semicircle)
- DAE = 180 – (90+35)=55^{\circ}

4. The points A, B, C, and D are points on the circumference of a circle. AC and BC intersect at the centre O at an angle of 152^{\circ} . Calculate the size of angle ABO .

76^{\circ}

28^{\circ}

14^{\circ}

90^{\circ}

- BOC = 152^{\circ} (vertically opposite angles are equal)
- OBC = (180-152)\div2 = 14^{\circ} (isosceles triangle)
- ABO = 90 – 14 = 76^{\circ}

5. The points A, B, C and D are on the circumference of a circle with centre O . The chords AC and BD intersect at the point G . AC is parallel to the tangent EF which passes through the point B . Angle FBC = 70^{\circ} . Calculate the size of angle BDC .

90^{\circ}

40^{\circ}

20^{\circ}

70^{\circ}

- CBG = 90 – 70 = 20^{\circ} (tangent meets radius at 90^{\circ} )
- BCD = 90^{\circ} (angles in a semicircle)
- BDC = 180 – (90+20) = 70^{\circ} (angles in a triangle)

6. The points A, B, C, D and E are on the circumference of a circle with centre O . AB = AE , and AD = AC. Angle ACD = 26^{\circ} . Calculate the size of angle DAE .

24^{\circ}

42^{\circ}

21^{\circ}

45^{\circ}

- CAD = 180 – (66+66) = 48^{\circ} (isosceles triangle)
- BAE = 90^{\circ} (angles in a semicircle)
- DAE = (90-48)\div2 = 21^{\circ}

1.

(a) The circle with centre O has a diameter AD . Points A, B, C, and D lie on the circumference of the circle and AC intersects with BD at the point E . Calculate the size of angle ABE marked x . Show your working.

(b) Hence calculate the size of angle BAE , marked y .

**(10 marks)**

Show answer

(a)

DCE = 90^{\circ}

**(1)**

The angle in a semicircle is 90^{\circ}

**(1)**

DEC = 180 – (90+40) = 50^{\circ}

**(1)**

Angles in a triangle

**(1)**

BEA = 50^{\circ}

**(1)**

Vertically opposite angles are equal

**(1)**

(b)

ABE = 90^{\circ}

**(1)**

The angle in a semicircle is 90^{\circ}

**(1)**

BAE = 180 – (90+50) = 40^{\circ}

**(1)**

Angles in a triangle

**(1)**

2. ABC is a triangle inscribed into a circle. Calculate the size of angle ABC .

**(4 marks)**

Show answer

BAC = 90^{\circ}

**(1)**

The angle in a semicircle is 90^{\circ}

**(1)**

ABC = 180 – (90 + 15) = 75^{\circ}

**(1)**

Angles in a triangle total 180^{\circ}

**(1)**

3.

(a) By calculating the size of angle ACD , find the size of angle \theta . State any reasons with your answer.

(b) What type of triangle is BCD ?

**(5 marks)**

Show answer

(a)

ACD = (180 – 120)\div2 = 30^{\circ}

**(1)**

ACB = 90^{\circ}

**(1)**

The angle in a semicircle is 90^{\circ}

**(1)**

BCD = 90 – 30 = 60^{\circ}

**(1)**

(b)

Equilateral

**(1)**

You have now learned how to:

- Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results

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#### GCSE Maths Papers - November 2022 Topics

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Practice paper packs based on the November advanced information for Edexcel 2022 Foundation and Higher exams.

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