Foundation June 2022 Paper 2R Q27
27 The diagram shows triangle \(ABP\) inside the regular hexagon \(ABCDEF\)
Diagram NOT accurately drawn
\(AB = 5\) cm \(BP = 2\) cm Angle \(ABP = 90°\)
Work out the size of angle \(PAF\)
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
eg \(\tan BAP = \dfrac{2}{5}\) or \(\sin BAP = \dfrac{2}{\sqrt{5^2 + 2^2}}\) or \(\dfrac{\sin BAP}{2} = \dfrac{\sin 90}{\sqrt{5^2 + 2^2}}\) \(\cos BAP = \dfrac{5}{\sqrt{5^2 + 2^2}}\) or \(\cos BAP = \dfrac{5^2 + (\sqrt{5^2 + 2^2})^2 - 2^2}{2 \times 5 \times \sqrt{29}}\) | M1 |
eg \((BAP =) \tan^{-1}\left(\dfrac{2}{5}\right)\) (= 21.8...) or \((BAP =) \sin^{-1}\left(\dfrac{2}{\sqrt{5^2 + 2^2}}\right)\) or \((BAP =) \sin^{-1}\left(\dfrac{2 \sin 90}{\sqrt{5^2 + 2^2}}\right)\) \((BAP =) \cos^{-1}\left(\dfrac{5}{\sqrt{5^2 + 2^2}}\right)\) or \(BAP = \cos^{-1}\left(\dfrac{5^2 + (\sqrt{5^2 + 2^2})^2 - 2^2}{2 \times 5 \times \sqrt{5^2 + 2^2}}\right)\) | M1 |
| eg (int angle =) \((6 - 2) \times 180 \div 6\) (= 120) or (ext angle =) \(360 \div 6\) (= 60) | M1 |
| eg “120” – “21.8” or 180 – “60” – “21.8” | M1 |
| 98.2 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for setting up a trig equation for angle \(BAP\)
M1: for a complete method to find angle \(BAP\) (= 21.8....)
[M2 for \(90 - \tan^{-1}\dfrac{5}{2}\) ie 90 – 68.2]
M1: Indep for a method to find the size of one interior or one exterior angle in a regular hexagon – could be seen on diagram
M1: for a complete method to find angle \(PAF\) where all values have come from a correct method
A1: accept 98.1 – 98.3