Higher June 2024 Paper 4 Q12
12
Angle SQR = 14° and angle SPQ = 62°.
Not to scale
Find the size of angle ROQ. [3]
Not to scale
Line GH is a tangent to the circle at F.
Line DE is parallel to line GH.
Complete these statements to prove that triangle DEF is isosceles.
Give reasons for your statements.
You may not need all of the lines.
Angle ............... = Angle ............... because ...............
Angle ............... = Angle ............... because ...............
Angle ............... = Angle ............... because ...............
Angle ............... = Angle ............... because ...............
Angle ............... = Angle ............... because ...............
Triangle DEF is isosceles because ............... [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 96 | 3 | B1 for angle SRQ = 180 – 62 or SRQ = 118 B1 for angle RSQ = 48 B1 for angle ROQ = 2 × their 48 to maximum of B2 | could be marked on the diagram, see appendix for alternative methods |
Appendix: working for Q12(a)
Hence angle ROQ = 96° from which ORQ=OQR=42°, SRO=76°, OQS=28°.
Alternative method 1
Some candidates are joining OP, however ROP are not co-linear but point P can be moved so that ROP is a straight line and maintaining angle SPQ as 62°.
Mark this;
B1 for angle SPO = 14°
B1 for angle OPQ and angle OQP = 48°
B1 for angle QOP = 84° or OQR and ORQ = 42°
to maximum of B2
Alternative method 2
They can draw a tangent at Q. T is the end of the tangent.
B1 for SQT = 62° (alternate segment theorem),
B1 for (angle OQT = 90°) so angle OQS = 28°,
B1 for making angle OQR = 42° = angle ORQ.
to maximum of B2
Hence angle ROQ = 180° – 42° – 42° = 96°
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Method 1 using angle EFH : angle DEF = angle EFH because alternate [angle] | B1 | Method 2 using angle DFG : angle EDF = angle DFG because alternate [angle] | If extra statements mark the best two BOD ‘alternating’ but not ‘Z-angles’ |
| angle EDF = angle EFH because alternate segment [thm.] | B1 | angle DEF = angle DFG because alternate segment [theorem] | BOD ‘alt. seg.’ |
| therefore angle DEF = angle EDF or “they have two angles equal” oe | B1dep | dep on B2 awarded If 0 scored SC1 for two sets of angles linked with incorrect/missing reasons e.g. DEF = EFH and EDF = EFH or EDF = DFG and DEF = DFG | Note : Angles may be written the other way round e.g. EDF is the same as FDE and they can say angle DFG = angle EDF and in marking you must use either method 1 or method 2 not both |