A, B and C are points on the circumference of a circle
O is the centre of the circle
angle OAB = 15°
angle BCO = 10°.
Not to scale
Calculate the acute angle AOC. [4]
(b) In the diagram,
E, F and D are points on the circumference of the circle
G, D and H lie on a tangent to the circle
angle EFD = 55°
angle FDE = 43°.
Not to scale
Explain why angle HDF is 82°. [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
50 nfww
4
M3 for ABC = 25 or B = 25 or for AOB = 150 and COB = 160 or M2 for ABO = 15 and CBO = 10 or for AOB = 150 or for COB = 160 or M1 for ABO = 15 or CBO = 10
If 0 scored, SC1 for AOC = 2 × [their] ABC stated or applied or for 360 – their AOB – their COB applied
Alternative method to find AOC = \(x\) M3 for \(\frac{x}{2} + 2\left(\frac{180 - x}{2}\right) + 15 + 10 = 180\) oe OR M1 for OAC = OCA = \(\frac{180 - x}{2}\) and M1 for ABC = \(\frac{x}{2}\)
Alternative method to find AOC = \(x\) M3 for \(360 - x + 10 + 15 + \frac{x}{2} = 360\) OR M1 for [reflex] AOC = \(360 - x\) and M1 for ABC = \(\frac{x}{2}\)
Throughout, angles could be on diagram
SC0 for angle at centre = 2 × angle at circumference
Mark scheme (b)
Answer
Marks
Part marks and guidance
e.g. DEF = 180 – (43 + 55) = 82 angles in a triangle HDF = DEF = 82 alternate segment theorem
OR
GDE = 55 alternate segment theorem HDF = 180 – (43+55) = 82 angles on a straight line
4
M2 for [DEF =] 180 – (43 + 55) soi by DEF = 82 and angles in a triangle or M1 for [DEF =] 180 – (43 + 55) soi by DEF = 82
AND
M2 for HDF = DEF [= 82] and alternate segment theorem or M1 for HDF = DEF [= 82]
Alternative method M2 for GDE = 55 and alternate segment theorem or M1 for GDE = 55
AND
M2 for [HDF =] 180 – (43 + 55) [= 82] and angles on a straight line or M1 for [HDF =] 180 – (43 + 55) [= 82]
Allow full marks if 3 letter angle notation not used provided their angles are unambiguously defined (eg. labelled on the diagram and referred to in working using their labels)
Note: 180 – (43 + 55) with no other creditable working or reasoning scores M1
