AB is a diameter of a circle with centre O. C is a point on the circumference. Work out the size of angle ACB. Give a brief reason.
(Total for Question 1 is 2 marks)
2
In a circle with centre O, angle AOB = 136 degrees. Point C lies on the circumference on the major arc AB. Work out angle ACB.
(Total for Question 2 is 2 marks)
3
Give a short proof that the angle at the centre is twice the angle at the circumference standing on the same arc. Use radii in your explanation.
(Total for Question 3 is 3 marks)
4
Points A, B, C and D lie on the same circle with A, B fixed and C and D on the same side of chord AB. Angle ACB = 37 degrees. Work out angle ADB. Give a reason.
(Total for Question 4 is 2 marks)
5
ABCD is a cyclic quadrilateral. Angle ABC = 68 degrees. Work out angle ADC and give a reason.
(Total for Question 5 is 3 marks)
6
Prove that opposite angles of a cyclic quadrilateral sum to 180 degrees. You may use the fact that the angle at the centre is twice the angle at the circumference.
(Total for Question 6 is 3 marks)
7
A line is tangent to a circle at A. B and C are points on the circumference so that B and C are distinct from A and lie on the circle. Show that the angle between the tangent at A and the chord AB is equal to the angle in the opposite arc at C, that is angle ACB.
(Total for Question 7 is 3 marks)
Mark scheme · 8.8D Proof of the Circle Theorems: Fluency and Exam Drill
Question 1
B1 90 degrees cao
B1 reason: angle in a semicircle is a right angle oe
Answer: 90 degrees; angle in a semicircle is a right angle.
Question 2
M1 use angle at centre = 2 x angle at circumference (or divide 136 by 2) oe
A1 68 degrees cao
Answer: 68 degrees
Question 3
M1 state OA = OC and OB = OC (radii), so triangles OAC and OBC are isosceles oe
M1 deduce angle OCA = angle OAC and angle OBC = angle OCB from the isosceles triangles oe
A1 angle AOB = 2 x angle ACB cso
Answer: Proof: OA = OC and OB = OC so triangles OAC and OBC are isosceles. Thus OCA = OAC and OBC = OCB. Adding these equal angles shows angle AOB = 2 x angle ACB. (cso)
Question 4
M1 identify angles ACB and ADB subtend the same arc AB (angles in same segment)
A1 37 degrees cao
Answer: 37 degrees; angles in the same segment are equal.
Question 5
M1 state opposite angles in a cyclic quadrilateral sum to 180 degrees oe
M1 calculate 180 - 68 oe
A1 112 degrees cao
Answer: 112 degrees; opposite angles in a cyclic quadrilateral sum to 180 degrees.
Question 6
M1 join the diagonals or consider the arc subtended by two opposite angles and use angle at centre twice circumference
M1 express each opposite angle as half of the same arc(s) so they add to half + half = whole
A1 sum = 180 degrees cso
Answer: Proof: Using angle at centre twice angle at circumference, the two opposite angles subtend the same full arc and so their halves add to the whole: sum = 180 degrees. (cso)
Question 7
M1 join OA and OB and use OA = OB (radii) so triangle OAB is isosceles and angle OAB = (180 - angle AOB)/2 oe
M1 use angle at centre = 2 x angle at circumference to write angle AOB = 2 x angle ACB and deduce angle OAB = 90 - angle ACB oe
A1 therefore angle between tangent at A and chord AB = angle ACB cso
Answer: Angle between tangent at A and chord AB = angle ACB.