Higher June 2023 Paper 4 Q17
17
(a) P, Q and R are points on the circumference of a circle, centre O.
Not to scale
Angle PRQ = 59° and angle PQR = 67°.
Line SPT is a tangent to the circle.
(i) Work out angle \(a\).
Give a reason for your answer. [2]
Give a reason for your answer. [2]
(ii) Work out angle \(b\).
Give a reason for your answer. [2]
Give a reason for your answer. [2]
(b) E, F, G and H are points on the circumference of a circle, centre O.
Acute angle EHG = \(x\)°.
Acute angle EHG = \(x\)°.
Not to scale
(i) Complete the following, giving the values of the angles in terms of \(x\).
Obtuse angle EOG = ........ ° because ....................
Therefore, reflex angle EOG = ........ °
Therefore, angle EFG = ........ °
[3](ii) Write down what your working in part (b)(i) has proved. [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (a)(i) | |||
| [\(a\)] 67° and [angles in] alternate segment [are equal] | 2 | B1 for 67 | Note : 0 for 59 (or any other value) and alternate segment Allow for reason : angle between chord and tangent equals angle in opposite segment |
| (a)(ii) | |||
| [\(b\)] 23° | 1 | FT 90 – their a (providing the answer is positive) | |
| [angle between] radius and tangent is 90° oe | 1 | Condone these terms : diameter, perpendicular, right-angle | |
Appendix: exemplar responses for Q17(a)(i)
| Response | Mark |
|---|---|
| Angles in the alternate segment are equal | 1 |
| Angles on a tangent which meet a triangle in a circle equal the alternate angle in the triangle | 0 |
Appendix: exemplar responses for Q17(a)(ii)
| Response | Mark |
|---|---|
| Angle between tangent and radius is 90 | 1 |
| Radius meets tangent at 90 | 1 |
| Tangent is perpendicular to radius | 1 |
| Tangent and diameter form a perpendicular bisector | 1 |
| Tangent subtends a 90 angle at the radius | 1 |
| Angles on a tangent on a radius are 90 | 1(BOD) |
| Line from centre to tangent at the point where the tangent touches the circle is 90° | 1(BOD) |
| Radius to line SPT will form a right angle | 0 |
| Angle between tangent and circle is 90 | 0 |
| Angle between tangent and a straight line is 90 | 0 |
| Angles at a right angle add up to 90 | 0 |
| Right angle between centre and tangent | 0 |
| Angles on a tangent add up to 90 | 0 |
| Tangent meets the chord at 90 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (b)(i) | |||
| \(2x\) [because] angle [subtended] at centre is twice angle at circumference oe | 1 | Condone “inscribed angle theorem” and “central angle theorem” | |
| \(360 - 2x\) | 1FT | STRICT FT e.g. 360 – their \(2x\) | For \(180 - x\) condone \(\frac{360 - 2x}{2}\) |
| \(180 - x\) | 1dep | Dep. on \(2x\) and \(360 - 2x\) | Do not accept working back from (b)(ii) unless whole part is complete and correct |
| (b)(ii) | |||
| Opposite angles [in a] cyclic quadrilateral [sum to] 180° | 1 | ||
Appendix: exemplar responses for Q17(b)(i)
| Response | Mark |
|---|---|
| Angles at centre is twice angle at circumference | 1 |
| Angles from the same chord are double at the centre than at the circumference | 1 |
| Angles on the perimeter are half the centre angle | 1 |
| Angle at the centre of a cyclic quadrilateral is twice the angle at the opposite point | 0 |
| Angles at centre is twice angle at the edge | 0 |
Appendix: exemplar responses for Q17(b)(ii)
| Response | Mark |
|---|---|
| Opposite angles in a cyclic quadrilateral add up to 180/supplementary | 1 |
| Opposite angles in a cyclic quadrilateral = 180 | 1(BOD) |
| Opposite angles in a quadrilateral add up to 180 | 0 |
| Angles in a [cyclic] quadrilateral add up to 360 | 0 |
| It is a cyclic quadrilateral | 0 |
| Cyclic quadrilateral theorem | 0 |