Higher November 2016 Unit 1 Q13
13. The points P, Q and R lie on the circumference of a circle, centre O.
PQ is a diameter of the circle.
The straight line ARB is a tangent to the circle.
\(Q\hat{R}B = x\), where \(x\) is measured in degrees.
Diagram not drawn to scale
Calculate the size of \(P\hat{Q}R\) in terms of \(x\).
You must give a reason for each step of your solution. [4]
| Scheme | Marks | Comments |
|---|---|---|
| RPQ = \(x\) | B1 | Check diagram. |
| (Reason =) Alternate segment (theorem) | E1 | Dependent on B1. |
| PQR = 90 – \(x\) | B1 | F.T. 90 – ‘their RPQ’. |
| (Reason =) Angle in a semicircle is 90° (and angles in a triangle add up to 180°) | E1 |
Accept correct alternative methods e.g.
| Scheme | Marks | Comments |
|---|---|---|
| (i) PRA = 90 – x | B1 | |
| (Reason =) Angle in a semicircle is 90° (and angles on a straight line add up to 180°) | E1 | |
| PQR =90 –x | B1 | FT PQR =’their PRA’ |
| (Reason =) Alternate segment (theorem) | E1 | |
| (ii) ORQ = 90 – x | B1 | |
| ‘radius and tangent meet at 90°’ | E1 | |
| PQR = 90 – x | B1 | |
| Isosceles triangle | E1 |