4MA0 Higher January 2018 Paper 3 Q25
25
Diagram NOT accurately drawn
The diagram shows a pyramid with horizontal base \(ABCDE\).
\(ABCDE\) is a regular pentagon, centre \(O\) and side 8 cm.
The vertex \(P\) is vertically above \(O\).
\(M\) is the midpoint of \(CD\).
\(OP = 10\) cm.
Calculate the size of angle \(APM\).
Give your answer correct to 1 decimal place.
(5)
| Scheme | Marks |
|---|---|
| e.g. angle \(AOB = 360 \div 5\) (=72) or angle \(ABC = 108\) or angle \(OBC = 54\) or angle \(COM = 36\) | M1 |
e.g. \(AO = \dfrac{4}{\cos(54)}\) (= 6.8(05...)) oe \(OM = 4\tan(54)\) (= 5.5(055...)) oe | M1 |
e.g. \(\tan APO = \dfrac{\text{“}6.8\ldots\text{”}}{10}\) (angle \(APO\) = 34(.2...)) e.g. \(\tan OPM = \dfrac{\text{“}5.5\ldots\text{”}}{10}\) (angle \(OPM\) = 28.8... or 29) | M1 |
| “34.2” + “28.8” OR 180 – “55.8” – “61.1” | M1 |
| 63.1 | A1 |
| (5) |
Notes
M1: for correct use of angles in a pentagon
M1: for a method to find \(AO\) (or \(BO\) etc) or \(OM\)
M1: for a method to find angle \(APO\) or angle \(OPM\) OR
for a method to find angle \(PAO\) (55.8...) AND angle \(PMO\) (61.1...)
M1: for a complete method
A1: for answer in the range 62.9 – 63.2
Alternative
| Scheme | Marks |
|---|---|
| e.g. angle \(AOB = 360 \div 5\) (=72) or angle \(ABC = 108\) or angle \(OBC = 54\) or angle \(COM = 36\) | M1 |
e.g. \(AO = \dfrac{4}{\cos(54)}\) (= 6.8(05...)) oe \(OM = 4\tan(54)\) (= 5.5(055...)) oe | M1 |
| e.g. \(AP = \sqrt{10^2 + 6.8..^2}\) (= 12.(09...)) AND \(MP = \sqrt{10^2 + 5.5..^2}\) (= 11.(415...)) | M1 |
| e.g. \(\cos APM = \dfrac{12.09..^2 + 11.415..^2 - 12.31..^2}{2 \times 12.09.. \times 11.415..}\) | M1 |
| 63.1 | A1 |
Notes
M1: for correct use of angles in a pentagon
M1: f for a method to find \(AO\) (or \(BO\) etc) or \(OM\)
M1: for a method to find \(AP\) (or \(BP\) etc) AND \(MP\)
M1: for a complete method
A1: for answer in the range 62.9 – 63.2