4MA0 Higher June 2015 Paper 3 Q15
15 \(ABCD\) is a trapezium.
Diagram NOT accurately drawn
\(AB = 25\) cm.
\(BC = 24\) cm.
\(CD = 10\) cm.
Angle \(ABC\) = angle \(BCD = 90^\circ\)
Calculate the size of angle \(CDA\).
Give your answer correct to 3 significant figures.
(4)
| Scheme | Marks |
|---|---|
| \(\tan x = \dfrac{25 - 10}{24}\) | M1 |
\((x =)\;\tan^{-1}\left(\dfrac{25 - 10}{24}\right)\) or \(\tan^{-1} 0.625\) or 32(.005…) | M1(dep) |
| 90 + “\(x\)” oe | M1 (indep) |
| 122 | A1 |
| (4) |
Notes
A1: awrt 122
Alternative
| Scheme | Marks |
|---|---|
| \(\tan A = \dfrac{24}{25 - 10}\) | M1 |
\((A =)\;\tan^{-1}\left(\dfrac{24}{25 - 10}\right)\) or \(\tan^{-1} 1.6\) or 58 or 57.9(94…) | M1(dep) |
| 360 – 90 – 90 – “\(A\)” oe | M1 (indep) |
| 122 | A1 |
Notes
A1: awrt 122
Alternative
| Scheme | Marks |
|---|---|
\((BDC =)\;\tan^{-1}\left(\dfrac{24}{10}\right)\) or \((BDC =)\) 67.4 or 67.3… | M1 |
| Fully correct method for \(BDA\) or \((ADB =)\) 54.6 | M1 |
| “54.6” + “67.4” | M1 (indep) |
| 122 | A1 |
Notes
M1: for a fully correct method to find angle BDC
M1: for a fully correct method to find angle \(BDA\)
A1: awrt 122