Higher November 2018 Paper 1 Q20
20 \(P\), \(Q\), \(R\) and \(S\) are points on a circle.
\(PXR\) and \(QXS\) are straight lines.
\(PX = SX\)
Not drawn accurately
Prove that \(QS\) is not a diameter of the circle. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| angle \(QPR = 27\) | M1 | may be seen on diagram |
| angle \(XPS = \dfrac{180 - 50}{2}\) or 65 | M1 | may be seen on diagram |
| angle \(QPR = 27\) and angle \(XPS = 65\) and angle \(QPS = 92\) and angle in a semicircle is a right angle | A1 | oe accept \(92 \ne 90\) |
| all reasons for angle facts: angles in same segment (are equal) and angle sum of triangle (is 180) and base angles of isosceles triangle (are equal) | A1 | oe |
| Alternative method 2 | ||
| angle \(SXR = 180 - 50\) or 130 and angle \(XRS = 180 -\) their \(130 - 27\) and angle \(PQS =\) their 23 | M1 | may be seen on diagram angle \(XRS = 23\) |
| angle \(XSP = \dfrac{180 - 50}{2}\) or 65 | M1 | may be seen on diagram |
| angle \(SXR = 130\) and angle \(XRS = 23\) and angle \(PQS = 23\) and \(XSP = 65\) and angle \(QPS = 92\) and angle in a semicircle is a right angle | A1 | oe accept \(92 \ne 90\) |
| all reasons for angle facts: angles on a straight line (add up to 180) and angle sum of triangle (is 180) and angles in same segment (are equal) and base angles of isosceles triangle (are equal) | A1 | oe |