Angle ACD = x° and angle OAB = 2x°.

Find an expression, in terms of x, in its simplest form
for

(a) angle AOB,

(b) angle ACB,

(c) angle DAB.

*(Cambridge Assessment International Education. 0580/22, October/November 2019, Q 19) *

(a) | |

(b) | |

(c) |

2. The diagram shows a circle with diameter PQ.

Find the value of y.

*(Cambridge Assessment International Education. 0580/42, May/June 2019, Q 2b) *

**3. A, B and C are points on the circle, center O.**

AB and OC intersect at P.

Find the value of w.

*(Cambridge Assessment International Education. 0580/22, February/March 2019, Q 15) *

4. In the diagram, A, B, C and D lie on the circle, centre O.

EA is a tangent to the circle at A.

Angle EAB = 61° and angle BAC = 55°.

(a) Find angle BAO.

(b) Find angle AOC.

(c) Find angle ABC.

(d) Find angle CDA.

*(Cambridge Assessment International Education. 0580/42, October/November 2018, Q 7)*

(a) | |

(b) | |

(c) | |

(d) |

5. A, B, C, D and E lie on the circle, centre O.

Angle AEB = 35°, angle ODE = 28° and angle ACD = 109°.

(i) Work out the following angles, giving reasons for your answers.

(a) Angle EBD

(b) Angle EAD

(ii) Work out angle BEO.

*(Cambridge Assessment International Education. 0580/42, May/June 2018, Q 9a)*

5(i)a | |

5(i)b | |

5(ii) |

6. A, B, C and D are points on the circle, center O. BCE is a straight line.

Angle AOC = 108° and angle DCE = 60°.

Calculate the values of w, x and y.

*(Cambridge Assessment International Education. 0580/22, October/November 2017, Q 22)*

7. A, B, C, D and E lie on the circle. AB is extended to F.

Angle AED = 140° and angle CBF = 95°.

Find the values of w, x and y.

*(Cambridge Assessment International Education. 0580/22, May/June 2017, Q 26)*

**8. The diagram shows points A, B, C and D on the circumference of a circle, centre O.AD is a straight line, AB = BC and angle OAB = 52°.**

Find angle ADC.

(c) The diagram shows points P, Q, R and S on the circumference of a circle, centre O.

VT is the tangent to the circle at Q.

Complete the statements.

(i) Angle QPS = angle QRS = ................ ° because ................

(ii) Angle SQP = ................ ° because .....................................

(iii) Part (c)(i) and part (c)(ii) show that

the cyclic quadrilateral PQRS is a ...........................................

*(Cambridge Assessment International Education. 0580/42, February/March 2017, Q 6b &6c)*

(c)(i) | Angle QPS = angle QRS = because 90°angle in a semi circle is a right angle |

(c)(ii) | Angle SQP = because 27°angle between a tangent and the radius drawn to thepoint of contact is 90° |

(c)(iii) | Part (c)(i) and part (c)(ii) show that the cyclic quadrilateral PQRS is a rectangle |