Higher June 2022 Paper 1 Q15
15.
Diagram not drawn to scale
\(P\), \(Q\), \(R\) and \(S\) are points on a circle with centre \(O\).
The line \(TU\) is a tangent to the circle at the point \(S\).
\(P\hat{O}S = 130^\circ\) and \(P\hat{Q}R = 120^\circ\).
Show that \(x = 55^\circ\).
Give a reason for each step of your answer. [5]
| Scheme | Marks | Comments |
|---|---|---|
| Angles may be marked on diagram. Degrees symbol may be omitted throughout. | ||
| \(P\hat{S}O = (180 - 130)\div2\) or \(25^\circ\) (base angle isosceles triangle) | B1 | |
| \(P\hat{S}R = 180 - 120\) or \(60^\circ\) (opposite angles in a cyclic quadrilateral) | B1 | |
| \(O\hat{S}R = 60 - 25\) oe or \(35^\circ\) \((P\hat{S}R = P\hat{S}O + O\hat{S}R)\) OR \(T\hat{S}P = 90 - 25\) or \(65^\circ\) (tangent perpendicular to radius) | B1 | FT ‘their derived 25’ and ‘their derived 60’ providing B1 already awarded; sight of 35 does not imply the previous mark |
| \(x = 90 - 35\) oe \((= 55^\circ)\) (tangent perpendicular to radius) OR \(x = 180 - 125\ (= 55^\circ)\) | B1 | CAO; must have earned all previous marks |
| At least ‘opposite angles in a cyclic quadrilateral’ oe and ‘tangent perpendicular to radius’ oe appropriately stated. | E1 | |
| If working back using the 55 award SC3 for \(90 - 55 = 35\) oe (tangent perpendicular to radius) \(P\hat{S}O = (180 - 130)\div2\) or \(25^\circ\) (base angle isosceles triangle) \(P\hat{S}R = 25 + 35\) or 60 \(P\hat{S}R = 180 - 120 = 60^\circ\) or \(P\hat{Q}R = 180 - 60 = 120\) (opposite angles in a cyclic quadrilateral) or SC2 for any two or three of the 4 correct statements or SC1 for any one correct statement and E1 for reasons as above | ||
| (5) |