Higher January 2023 Paper 2R Q23
23 \(AEC\) and \(BED\) are chords of a circle.
Diagram NOT accurately drawn
\(AE = (x + 5)\) cm \(BE = x\) cm \(CE = (5x - 12)\) cm \(DE = (x + 12)\) cm
Angle \(DAE = 48°\)
Work out the size of angle \(ADE\)
Give your answer correct to one decimal place.
(5)
| Scheme | Marks |
|---|---|
| eg \((x + 5)(5x - 12) = x(x + 12)\) | M1 |
| eg \(4x^2 + x - 60\;(= 0)\) oe allow \(4x^2 + x = 60\) | A1 |
eg \((4x - 15)(x + 4)\;(= 0)\) or \(\dfrac{-1 \pm \sqrt{1^2 - 4 \times 4 \times -60}}{2 \times 4}\) or \(4\left[\left(x + \dfrac{1}{8}\right)^2 - \left(\dfrac{1}{8}\right)^2\right] = 60\) oe | M1 |
| eg \((ADE =)\;\sin^{-1}\left(\dfrac{(\text{``}{3.75}\text{''} + 5)\sin(48)}{\text{``}{3.75}\text{''} + 12}\right)\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 24.4 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for setting up a correct equation
A1: for writing the correct quadratic expression in the form \(ax^2 + bx + c\;(= 0)\)
allow \(ax^2 + bx = c\)
M1: (dep on M1) for a complete method to solve their 3-term quadratic (allow one sign error and some simplification – allow as far as \(\dfrac{-1 \pm \sqrt{1 + 960}}{8}\))
Allow + instead of ± in quadratic formula
M1: for a complete method for \(ADE\). Allow use of \(x = -4\) for this mark
A1: accept 24.3 – 24.4